Derivatives. By: OpenStaxCollege

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1 By: OpenStaxCollege Te average teen in te United States opens a refrigerator door an estimated 25 times per day. Supposedly, tis average is up from 10 years ago wen te average teenager opened a refrigerator door 20 times per day ttp:// Source provided.. It is estimated tat a television is on in a ome 6.75 ours per day, wereas parents spend an estimated 5.5 minutes per day aving a meaningful conversation wit teir cildren. Tese averages, too, are not te same as tey were 10 years ago, wen te television was on an estimated 6 ours per day in te typical ouseold, and parents spent 12 minutes per day in meaningful conversation wit teir kids. Wat do tese scenarios ave in common? Te functions representing tem ave canged over time. In tis section, we will consider metods of computing suc canges over time. Finding te Average Rate of Cange of a Function Te functions describing te examples above involve a cange over time. Cange divided by time is one example of a rate. Te rates of cange in te previous examples are eac different. In oter words, some canged faster tan oters. If we were to grap te functions, we could compare te rates by determining te slopes of te graps. A tangent line to a curve is a line tat intersects te curve at only a single point but does not cross it tere. (Te tangent line may intersect te curve at anoter point away from te point of interest.) If we zoom in on a curve at tat point, te curve appears linear, and te slope of te curve at tat point is close to te slope of te tangent line at tat point. [link] represents te function f(x) = x 3 4x. We can see te slope at various points along te curve. slope at x = 2 is 8 1/53

2 slope at x = 1 is 1 slope at x = 2 is 8 Grap sowing tangents to curve at 2, 1, and 2. Let s imagine a point on te curve of function f at x = a as sown in [link]. Te coordinates of te point are (a, f(a)). Connect tis point wit a second point on te curve a little to te rigt of x = a, wit an x-value increased by some small real number. Te coordinates of tis second point are (a +, f(a + )) for some positive-value. 2/53

3 Connecting point a wit a point just beyond allows us to measure a slope close to tat of a tangent line at x = a. We can calculate te slope of te line connecting te two points (a, f(a)) and (a +, f(a + )), called a secant line, by applying te slope formula, slope = slope = cange in y cange in x cange in y cange in x We use te notation m sec to represent te slope of te secant line connecting two points. m sec = = f(a + ) f(a) (a + ) (a) f(a + ) f(a) a + a Te slope m sec equals te average rate of cange between two points (a, f(a)) and (a +, f(a + )). m sec = f(a + ) f(a) A General Note Te Average Rate of Cange between Two Points on a Curve Te average rate of cange (AROC) between two points (a, f(a)) and (a +, f(a + )) on te curve of f is te slope of te line connecting te two points and is given by 3/53

4 f(a + ) f(a) AROC = Finding te Average Rate of Cange Find te average rate of cange connecting te points (2, 6) and ( 1, 5). We know te average rate of cange connecting two points may be given by AROC = f(a + ) f(a). If one point is (2, 6), or (2, f(2)), ten f(2) = 6. Te value is te displacement from 2 to 1, wic equals 1 2 = 3. For te oter point, f(a + ) is te y-coordinate at a +, wic is 2 + ( 3) or 1, so f(a + ) = f( 1) = 5. AROC = Try It f(a + ) f(a) = 5 ( 6) 3 = 11 3 = 11 3 Find te average rate of cange connecting te points ( 5, 1.5) and ( 2.5, 9). 3 Understanding te Instantaneous Rate of Cange Now tat we can find te average rate of cange, suppose we make in [link] smaller and smaller. Ten a + will approac a as gets smaller, getting closer and closer to 0. Likewise, te second point (a +, f(a + )) will approac te first point, (a, f(a)). As a consequence, te connecting line between te two points, called te secant line, will get closer and closer to being a tangent to te function at x = a, and te slope of te secant line will get closer and closer to te slope of te tangent at x = a. See [link]. 4/53

5 Te connecting line between two points moves closer to being a tangent line at x = a. Because we are looking for te slope of te tangent at x = a, we can tink of te measure of te slope of te curve of a function f at a given point as te rate of cange at a particular instant. We call tis slope te instantaneous rate of cange, or te derivative of te function at x = a. Bot can be found by finding te limit of te slope of a line connecting te point at x = a wit a second point infinitesimally close along te curve. For a function f bot te instantaneous rate of cange of te function and te derivative of te function at x = a are written as f'(a), and we can define tem as a twosided limit tat as te same value weter approaced from te left or te rigt. f f(a + ) f(a) (a) 0 Te expression by wic te limit is found is known as te difference quotient. A General Note Definition of Instantaneous Rate of Cange and Derivative Te derivative, or instantaneous rate of cange, of a function f at x = a, is given by f(a + ) f(a) f'(a) 0 5/53

6 f(a + ) f(a) Te expression is called te difference quotient. We use te difference quotient to evaluate te limit of te rate of cange of te function as approaces 0. : Interpretations and Notation Te derivative of a function can be interpreted in different ways. It can be observed as te beavior of a grap of te function or calculated as a numerical rate of cange of te function. Te derivative of a function f(x) at a point x = a is te slope of te tangent line to te curve f(x) at x = a. Te derivative of f(x) at x = a is written f (a). Te derivative f (a) measures ow te curve canges at te point (a, f(a)). Te derivative f (a) may be tougt of as te instantaneous rate of cange of te function f(x) at x = a. If a function measures distance as a function of time, ten te derivative measures te instantaneous velocity at time t = a. A General Note Notations for te Derivative Te equation of te derivative of a function f(x) is written as y = f (x), were y = f(x). Te notation f (x) is read as f prime of x. Alternate notations for te derivative include te following: f (x) = y = dy dx = df dx = d dxf(x) = Df(x) Te expression f (x) is now a function of x; tis function gives te slope of te curve y = f(x) at any value of x. Te derivative of a function f(x) at a point x = a is denoted f (a). How To Given a function f, find te derivative by applying te definition of te derivative. 1. Calculate f(a + ). 2. Calculate f(a). 3. Substitute and simplify f(a + ) f(a). 4. Evaluate te limit if it exists: f (a) 0 Finding te Derivative of a Polynomial Function f(a + ) f(a). 6/53

7 Find te derivative of te function f(x) = x 2 3x + 5 at x = a. We ave: f f(a + ) f(a) (a) 0 Definition of a derivative Substitute f(a + ) = (a + ) 2 3(a + ) + 5 and f(a) = a 2 3a + 5. f (a + )(a + ) 3(a + ) + 5 (a (a) 2 3a + 5) 0 Try It a 2 + 2a + 2 3a a 2 + 3a 5 0 a 2 + 2a + 2 3a 3+5 a 2 +3a 5 0 2a (2a + 3) 0 = 2a = 2a 3 Evaluate to remove parenteses. Simplify. Factor out an. Evaluate te limit. Find te derivative of te function f(x) = 3x 2 + 7x at x = a. f (a) = 6a + 7 Finding of Rational Functions To find te derivative of a rational function, we will sometimes simplify te expression using algebraic tecniques we ave already learned. Finding te Derivative of a Rational Function Find te derivative of te function f(x) = 3 + x 2 x at x = a. 7/53

8 f f(a + ) f(a) (a) (a + ) 0 2 (a + ) ( 3 + a 2 a) (2 (a + ))(2 a) [ 3 + (a + ) 0 2 a)] 2 (a + ) ( 3 + a (2 (a + ))(2 a)() (2 (a + ))(2 a) ( 3 + (a + ) 0 2 a) (2 (a + ))) (2 (a + ))(2 a) ( 3 + a (2 (a + ))(2 a)() 6 3a + 2a a a 6 + 3a + 3 2a + a 2 + a (2 (a + ))(2 a)() 0 5 (2 (a + ))(2 a)() 0 5 (2 (a + ))(2 a) 0 = 5 (2 (a + 0))(2 a) = 5 (2 a)(2 a) = 5 (2 a) 2 Substitute f(a + ) and f(a) Multiply numerator and denominator by (2 Distribute Multiply Combine like terms Cancel like factors Evaluate te limit Try It Find te derivative of te function f(x) = 10x x + 4 at x = a. f (a) = 15 (5a + 4) 2 Finding of Functions wit Roots To find derivatives of functions wit roots, we use te metods we ave learned to find limits of functions wit roots, including multiplying by a conjugate. Finding te Derivative of a Function wit a Root Find te derivative of te function f(x) = 4 x at x = 36. We ave f f(a + ) f(a) (a) 0 4 a + 4 a 0 Substitute f(a + ) and f(a) 8/53

9 4 a a Multiply te numerator and denominator by te conjugate: 4 a a. f (a) 0( 4 a + 4 a 0( 0( 0( 0( = 16 8 a = 2 a f (36) = (a + ) 16a 4( a a)) 16a a 4( a a)) 16 (4 a a)) 16 4 a a) ) ( 4 a a 4 a a) Multiply. Distribute and combine like terms. Simplify. Evaluate te limit by letting = 0. Evaluate te derivative at x = 36. = 2 6 = 1 3 Try It Find te derivative of te function f(x) = 9 x at x = Finding Instantaneous Rates of Cange Many applications of te derivative involve determining te rate of cange at a given instant of a function wit te independent variable time wic is wy te term instantaneous is used. Consider te eigt of a ball tossed upward wit an initial velocity of 64 feet per second, given by s(t) = 16t t + 6, were t is measured in seconds and s(t) is measured in feet. We know te pat is tat of a parabola. Te derivative will tell us ow te eigt is canging at any given point in time. Te eigt of te ball is sown in [link] as a function of time. In pysics, we call tis te s-t grap. 9/53

10 Finding te Instantaneous Rate of Cange Using te function above, s(t) = 16t t + 6, wat is te instantaneous velocity of te ball at 1 second and 3 seconds into its fligt? Te velocity at t = 1 and t = 3 is te instantaneous rate of cange of distance per time, or velocity. Notice tat te initial eigt is 6 feet. To find te instantaneous velocity, we find te derivative and evaluate it at t = 1 and t = 3 : 10/53

11 f f(a + ) f(a) (a) (t + ) (t + ) + 6 ( 16t t + 6) Substitute s(t + ) and s(t) t 2 32t t t 2 64t 6 32t ( 32t + 64) 0 32t s (t) = 32t + 64 Distribute. Simplify. Factor te numerator. Cancel out te common factor. Evaluate te limit by letting = 0 For any value of t, s (t) tells us te velocity at tat value of t. Evaluate t = 1 and t = 3. s (1) = 32(1) + 64 = 32 s (3) = 32(3) + 64 = 32 Te velocity of te ball after 1 second is 32 feet per second, as it is on te way up. Te velocity of te ball after 3 seconds is 32 feet per second, as it is on te way down. Try It Te position of te ball is given by s(t) = 16t t + 6. Wat is its velocity 2 seconds into fligt? 0 Using Graps to Find Instantaneous Rates of Cange We can estimate an instantaneous rate of cange at x = a by observing te slope of te curve of te function f(x) at x = a. We do tis by drawing a line tangent to te function at x = a and finding its slope. How To 11/53

12 Given a grap of a function f(x), find te instantaneous rate of cange of te function at x = a. 1. Locate x = a on te grap of te function f(x). 2. Draw a tangent line, a line tat goes troug x = a at a and at no oter point in tat section of te curve. Extend te line far enoug to calculate its slope as cange in y cange in x. Estimating te Derivative at a Point on te Grap of a Function From te grap of te function y = f(x) presented in [link], estimate eac of te following: 1. f(0) 2. f(2) 3. f'(0) 4. f'(2) To find te functional value, f(a), find te y-coordinate at x = a. 12/53

13 To find te derivative at x = a, f (a), draw a tangent line at x = a, and estimate te slope of tat tangent line. See [link]. 1. f(0) is te y-coordinate at x = 0. Te point as coordinates (0, 1), tus f(0) = f(2) is te y-coordinate at x = 2. Te point as coordinates (2, 1), tus f(2) = f (0) is found by estimating te slope of te tangent line to te curve at x = 0. Te tangent line to te curve at x = 0 appears orizontal. Horizontal lines ave a slope of 0, tus f (0) = f (2) is found by estimating te slope of te tangent line to te curve at x = 2. Observe te pat of te tangent line to te curve at x = 2. As te x value moves one unit to te rigt, te y value moves up four units to anoter point on te line. Tus, te slope is 4, so f (2) = 4. Try It Using te grap of te function f(x) = x 3 3x sown in [link], estimate: f(1), f (1), f(0), and f (0). 13/53

14 2, 0, 0, 3 Using Instantaneous Rates of Cange to Solve Real-World Problems Anoter way to interpret an instantaneous rate of cange at x = a is to observe te function in a real-world context. Te unit for te derivative of a function f(x) is output units input unit Suc a unit sows by ow many units te output canges for eac one-unit cange of input. Te instantaneous rate of cange at a given instant sows te same ting: te units of cange of output per one-unit cange of input. One example of an instantaneous rate of cange is a marginal cost. For example, suppose te production cost for a company to produce x items is given by C(x), in tousands of dollars. Te derivative function tells us ow te cost is canging for 14/53

15 any value of x in te domain of te function. In oter words, C (x) is interpreted as a marginal cost, te additional cost in tousands of dollars of producing one more item wen x items ave been produced. For example, C (11) is te approximate additional cost in tousands of dollars of producing te 12 t item after 11 items ave been produced.c (11) = 2.50 means tat wen 11 items ave been produced, producing te 12 t item would increase te total cost by approximately $2, Finding a Marginal Cost Te cost in dollars of producing x laptop computers in dollars is f(x) = x 2 100x. At te point were 200 computers ave been produced, wat is te approximate cost of producing te 201 st unit? If f(x) = x 2 100x describes te cost of producing x computers, f (x) will describe te marginal cost. We need to find te derivative. For purposes of calculating te derivative, we can use te following functions: f(a + b) = (x + ) 2 100(x + ) f(a) = a 2 100a f (x) = f(a + ) f(a) Formula for a derivative = (x + )2 100(x + ) (x 2 100x) = x2 + 2x x 100 x x = 2x = (2x + 100) = 2x = 2x 100 f (x) = 2x 100 f (200) = 2(200) 100 = 300 Substitute f(a + ) and f(a). Multiply polynomials, distribute. Collect like terms. Factor and cancel like terms. Simplify. Evaluate wen = 0. Formula for marginal cost Evaluate for 200 units. Te marginal cost of producing te 201 st unit will be approximately $300. Interpreting a Derivative in Context 15/53

16 A car leaves an intersection. Te distance it travels in miles is given by te function f(t), were t represents ours. Explain te following notations: 1. f(0) = 0 2. f (1) = f(1) = f(2.5) = 150 First we need to evaluate te function f(t) and te derivative of te function f (t), and distinguis between te two. Wen we evaluate te function f(t), we are finding te distance te car as traveled in t ours. Wen we evaluate te derivative f (t), we are finding te speed of te car after t ours. 1. f(0) = 0 means tat in zero ours, te car as traveled zero miles. 2. f (1) = 60 means tat one our into te trip, te car is traveling 60 miles per our. 3. f(1) = 70 means tat one our into te trip, te car as traveled 70 miles. At some point during te first our, ten, te car must ave been traveling faster tan it was at te 1-our mark. 4. f(2.5) = 150 means tat two ours and tirty minutes into te trip, te car as traveled 150 miles. Try It A runner runs along a straigt east-west road. Te function f(t) gives ow many feet eastward of er starting point se is after t seconds. Interpret eac of te following as it relates to te runner. 1. f(0) = 0 2. f(10) = f (10) = f (20) = f(40) = After zero seconds, se as traveled 0 feet. 2. After 10 seconds, se as traveled 150 feet east. 3. After 10 seconds, se is moving eastward at a rate of 15 ft/sec. 4. After 20 seconds, se is moving westward at a rate of 10 ft/sec. 5. After 40 seconds, se is 100 feet westward of er starting point. 16/53

17 Finding Points Were a Function s Derivative Does Not Exist To understand were a function s derivative does not exist, we need to recall wat normally appens wen a function f(x) as a derivative at x = a. Suppose we use a graping utility to zoom in on x = a. If te function f(x) is differentiable, tat is, if it is a function tat can be differentiated, ten te closer one zooms in, te more closely te grap approaces a straigt line. Tis caracteristic is called linearity. Look at te grap in [link]. Te closer we zoom in on te point, te more linear te curve appears. We migt presume te same ting would appen wit any continuous function, but tat is not so. Te function f(x) = x, for example, is continuous at x = 0, but not differentiable at x = 0. As we zoom in close to 0 in [link], te grap does not approac a straigt line. No matter ow close we zoom in, te grap maintains its sarp corner. 17/53

18 Grap of te function f(x) = x, wit x-axis from 0.1 to 0.1 and y-axis from 0.1 to 0.1. We zoom in closer by narrowing te range to produce [link] and continue to observe te same sape. Tis grap does not appear linear at x = 0. Grap of te function f(x) = x, wit x-axis from to and y-axis from to /53

19 Wat are te caracteristics of a grap tat is not differentiable at a point? Here are some examples in wic function f(x) is not differentiable at x = a. In [link], we see te grap of f(x) = { x2, 8 x, x 2 x > 2. Notice tat, as x approaces 2 from te left, te left-and limit may be observed to be 4, wile as x approaces 2 from te rigt, te rigt-and limit may be observed to be 6. We see tat it as a discontinuity at x = 2. Te grap of f(x) as a discontinuity at x = 2. In [link], we see te grap of f(x) = x. We see tat te grap as a corner point at x = 0. 19/53

20 Te grap of f(x) = x as a corner point at x = 0. In [link], we see tat te grap of f(x) = x 2 3 as a cusp at x = 0. A cusp as a unique feature. Moving away from te cusp, bot te left-and and rigt-and limits approac eiter infinity or negative infinity. Notice te tangent lines as x approaces 0 from bot te left and te rigt appear to get increasingly steeper, but one as a negative slope, te oter as a positive slope. Te grap of f(x) = x 2 3 as a cusp at x = 0. In [link], we see tat te grap of f(x) = x 1 3 as a vertical tangent at x = 0. Recall tat vertical tangents are vertical lines, so were a vertical tangent exists, te slope of te line is undefined. Tis is wy te derivative, wic measures te slope, does not exist tere. 20/53

21 A General Note Differentiability Te grap of f(x) = x 1 3 as a vertical tangent at x = 0. A function f(x) is differentiable at x = a if te derivative exists at x = a, wic means tat f (a) exists. Tere are four cases for wic a function f(x) is not differentiable at a point x = a. 1. Wen tere is a discontinuity at x = a. 2. Wen tere is a corner point at x = a. 3. Wen tere is a cusp at x = a. 4. Any oter time wen tere is a vertical tangent at x = a. Determining Were a Function Is Continuous and Differentiable from a Grap Using [link], determine were te function is 1. continuous 2. discontinuous 3. differentiable 4. not differentiable At te points were te grap is discontinuous or not differentiable, state wy. 21/53

22 Te grap of f(x) is continuous on (, 2) ( 2, 1) (1, ). Te grap of f(x) as a removable discontinuity at x = 2 and a jump discontinuity at x = 1. See [link]. 22/53

23 Tree intervals were te function is continuous Te grap of is differentiable on (, 2) ( 2, 1) ( 1,1) (1,2) (2, ). Te grap of f(x) is not differentiable at x = 2 because it is a point of discontinuity, at x = 1 because of a sarp corner, at x = 1 because it is a point of discontinuity, and at x = 2 because of a sarp corner. See [link]. 23/53

24 Try It Five intervals were te function is differentiable Determine were te function y = f(x) sown in [link] is continuous and differentiable from te grap. 24/53

25 Te grap of f is continuous on (, 1) (1, 3) (3, ). Te grap of f is discontinuous at x = 1 and x = 3. Te grap of f is differentiable on (, 1) (1, 3) (3, ). Te grap of f is not differentiable at x = 1 and x = 3. Finding an Equation of a Line Tangent to te Grap of a Function Te equation of a tangent line to a curve of te function f(x) at x = a is derived from te point-slope form of a line, y = m(x x 1) + y 1. Te slope of te line is te slope of te curve at x = a and is terefore equal to f (a), te derivative of f(x) at x = a. Te coordinate pair of te point on te line at x = a is (a, f(a)). If we substitute into te point-slope form, we ave 25/53

26 Te equation of te tangent line is y = f'(a)(x a) + f(a) A General Note Te Equation of a Line Tangent to a Curve of te Function f Te equation of a line tangent to te curve of a function f at a point x = a is y = f'(a)(x a) + f(a) How To Given a function f, find te equation of a line tangent to te function at x = a. 1. Find te derivative of f(x) at x = a using f f(a + ) f(a) (a) Evaluate te function at x = a. Tis is f(a). 3. Substitute (a, f(a)) and f (a) into y = f'(a)(x a) + f(a). 4. Write te equation of te tangent line in te form y = mx + b. Finding te Equation of a Line Tangent to a Function at a Point Find te equation of a line tangent to te curve f(x) = x 2 4x at x = 3. Using: f(a + ) f(a) f'(a) 0 Substitute f(a + ) = (a + ) 2 4(a + ) and f(a) = a 2 4a. 26/53

27 f (a + )(a + ) 4(a + ) (a 2 4a) (a) 0 a 2 + 2a + 2 4a 4 a 2 + 4a 0 a 2 + 2a + 2 4a 4 a 2 +4a 0 2a (2a + 4) 0 = 2a f (a) = 2a 4 f (3) = 2(3) 4 = 2 Remove parenteses. Combine like terms. Factor out. Evaluate te limit. Equation of tangent line at x = 3: y = f'(a)(x a) + f(a) y = f'(3)(x 3) + f(3) y = 2(x 3) + ( 3) y = 2x 9 Analysis We can use a graping utility to grap te function and te tangent line. In so doing, we can observe te point of tangency at x = 3 as sown in [link]. 27/53

28 Try It Grap confirms te point of tangency at x = 3. Find te equation of a tangent line to te curve of te function f(x) = 5x 2 x + 4 at x = 2. y = 19x 16 Finding te Instantaneous Speed of a Particle If a function measures position versus time, te derivative measures displacement versus time, or te speed of te object. A cange in speed or direction relative to a cange in time is known as velocity. Te velocity at a given instant is known as instantaneous velocity. In trying to find te speed or velocity of an object at a given instant, we seem to encounter a contradiction. We normally define speed as te distance traveled divided by te elapsed time. But in an instant, no distance is traveled, and no time elapses. How will we divide zero by zero? Te use of a derivative solves tis problem. A derivative allows us to say tat even wile te object s velocity is constantly canging, it as a certain velocity at a given instant. Tat means tat if te object traveled at tat exact velocity for a unit of time, it would travel te specified distance. A General Note 28/53

29 Instantaneous Velocity Let te function s(t) represent te position of an object at time t. Te instantaneous velocity or velocity of te object at time t = a is given by s s(a + ) s(a) (a) 0 Finding te Instantaneous Velocity A ball is tossed upward from a eigt of 200 feet wit an initial velocity of 36 ft/sec. If te eigt of te ball in feet after t seconds is given by s(t) = 16t t + 200, find te instantaneous velocity of te ball at t = 2. First, we must find te derivative s (t). Ten we evaluate te derivative at t = 2, using s(a + ) = 16(a + ) (a + ) and s(a) = 16a a s s(a + ) s(a) (a) 0 16(a + ) (a + ) ( 16a a + 200) 0 16(a 2 + 2a + 2 ) + 36(a + ) ( 16a a + 200) 0 16a 2 32a a a 2 36a a 2 32a a a 2 36a a ( 32a ) 0 ( 32a ) 0 = 32a s (a) = 32a + 36 s (2) = 32(2) + 36 = 28 Analysis Tis result means tat at time t = 2 seconds, te ball is dropping at a rate of 28 ft/sec. 29/53

30 Try It A fireworks rocket is sot upward out of a pit 12 ft below te ground at a velocity of 60 ft/sec. Its eigt in feet after t seconds is given by s = 16t t 12. Wat is its instantaneous velocity after 4 seconds? 68 ft/sec, it is dropping back to Eart at a rate of 68 ft/s. Media Access tese online resources for additional instruction and practice wit derivatives. Estimate te Derivative Estimate te Derivative Ex. 4 Visit tis website for additional practice questions from Learningpod. Key Equations average rate of cange AROC = f(a + ) f(a) derivative of a function f f(a + ) f(a) (a) 0 Key Concepts Te slope of te secant line connecting two points is te average rate of cange of te function between tose points. See [link]. Te derivative, or instantaneous rate of cange, is a measure of te slope of te curve of a function at a given point, or te slope of te line tangent to te curve at tat point. See [link], [link], and [link]. Te difference quotient is te quotient in te formula for te instantaneous rate of cange: f(a + ) f(a) Instantaneous rates of cange can be used to find solutions to many real-world problems. See [link]. Te instantaneous rate of cange can be found by observing te slope of a function at a point on a grap by drawing a line tangent to te function at tat point. See [link]. Instantaneous rates of cange can be interpreted to describe real-world situations. See [link] and [link]. Some functions are not differentiable at a point or points. See [link]. 30/53

31 Te point-slope form of a line can be used to find te equation of a line tangent to te curve of a function. See [link]. Velocity is a cange in position relative to time. Instantaneous velocity describes te velocity of an object at a given instant. Average velocity describes te velocity maintained over an interval of time. Using te derivative makes it possible to calculate instantaneous velocity even toug tere is no elapsed time. See [link]. Section Exercises Verbal How is te slope of a linear function similar to te derivative? Te slope of a linear function stays te same. Te derivative of a general function varies according to x. Bot te slope of a line and te derivative at a point measure te rate of cange of te function. Wat is te difference between te average rate of cange of a function on te interval [x, x + ] and te derivative of te function at x? A car traveled 110 miles during te time period from 2:00 P.M. to 4:00 P.M. Wat was te car's average velocity? At exactly 2:30 P.M., te speed of te car registered exactly 62 miles per our. Wat is anoter name for te speed of te car at 2:30 P.M.? Wy does tis speed differ from te average velocity? Average velocity is 55 miles per our. Te instantaneous velocity at 2:30 p.m. is 62 miles per our. Te instantaneous velocity measures te velocity of te car at an instant of time wereas te average velocity gives te velocity of te car over an interval. Explain te concept of te slope of a curve at point x. Suppose water is flowing into a tank at an average rate of 45 gallons per minute. Translate tis statement into te language of matematics. Te average rate of cange of te amount of water in te tank is 45 gallons per minute. If f(x) is te function giving te amount of water in te tank at any time t, ten te average rate of cange of f(x) between t = a and t = b is f(a) + 45(b a). Algebraic f(x + ) f(x) For te following exercises, use te definition of derivative lim to calculate 0 te derivative of eac function. 31/53

32 f(x) = 3x 4 f(x) = 2x + 1 f (x) = 2 f(x) = x 2 2x + 1 f(x) = 2x 2 + x 3 f (x) = 4x + 1 f(x) = 2x f(x) = 1 x 2 f (x) = f(x) = 2 + x 1 x 1 (x 2) 2 f(x) = 5 2x 3 + 2x 16 (3 + 2x) 2 f(x) = 1 + 3x f(x) = 3x 3 x 2 + 2x + 5 f (x) = 9x 2 2x + 2 f(x) = 5 f(x) = 5π f (x) = 0 For te following exercises, find te average rate of cange between te two points. ( 2, 0) and ( 4, 5) (4, 3) and ( 2, 1) 32/53

33 1 3 (0, 5) and (6, 5) (7, 2) and (7, 10) undefined For te following polynomial functions, find te derivatives. f(x) = x f(x) = 3x 2 7x = 6 f (x) = 6x 7 f(x) = 7x 2 f(x) = 3x 3 + 2x 2 + x 26 f (x) = 9x 2 + 4x + 1 For te following functions, find te equation of te tangent line to te curve at te given point x on te curve. f(x) = 2x 2 3x x = 3 f(x) = x x = 2 y = 12x 15 f(x) = x x = 9 For te following exercise, find k suc tat te given line is tangent to te grap of te function. f(x) = x 2 kx, y = 4x 9 k = 10 or k = 2 33/53

34 Grapical For te following exercises, consider te grap of te function f and determine were te function is continuous/discontinuous and differentiable/not differentiable. 34/53

35 Discontinuous at x = 2 and x = 0. Not differentiable at 2, 0, 2. 35/53

36 36/53

37 Discontinuous at x = 5. Not differentiable at -4, 2, 0, 1, 3, 4, 5. For te following exercises, use [link] to estimate eiter te function at a given value of x or te derivative at a given value of x, as indicated. 37/53

38 f( 1) f(0) f(0) = 2 f(1) f(2) f(2) = 6 38/53

39 f(3) f ( 1) f ( 1) = 9 f (0) f (1) f (1) = 3 f (2) f (3) f (3) = 9 Sketc te function based on te information below: f (x) = 2x, f(2) = 4 Tecnology Numerically evaluate te derivative. Explore te beavior of te grap of f(x) = x 2 around x = 1 by graping te function on te following domains: [0.9, 1.1], [0.99, 1.01], [0.999, 1.001], and [0.9999, ]. We can use te feature on our calculator tat automatically sets Ymin and Ymax to te Xmin and Xmax values we preset. (On some of te commonly used graping calculators, tis feature may be called ZOOM FIT or ZOOM AUTO). By examining te corresponding range values for tis viewing window, approximate ow te curve canges at x = 1, tat is, approximate te derivative at x = 1. Answers vary. Te slope of te tangent line near x = 1 is 2. Real-World Applications For te following exercises, explain te notation in words. Te volume f(t) of a tank of gasoline, in gallons, t minutes after noon. f(0) = 600 f'(30) = 20 39/53

40 At 12:30 p.m., te rate of cange of te number of gallons in te tank is 20 gallons per minute. Tat is, te tank is losing 20 gallons per minute. f(30) = 0 f'(200) = 30 At 200 minutes after noon, te volume of gallons in te tank is canging at te rate of 30 gallons per minute. f(240) = 500 For te following exercises, explain te functions in words. Te eigt, s, of a projectile after t seconds is given by s(t) = 16t t. s(2) = 96 Te eigt of te projectile after 2 seconds is 96 feet. s'(2) = 16 s(3) = 96 Te eigt of te projectile at t = 3 seconds is 96 feet. s'(3) = 16 s(0) = 0, s(5) = 0. Te eigt of te projectile is zero at t = 0 and again at t = 5. In oter words, te projectile starts on te ground and falls to eart again after 5 seconds. For te following exercises, te volume V of a spere wit respect to its radius r is given by V = 4 3 πr3. Find te average rate of cange of V as r canges from 1 cm to 2 cm. Find te instantaneous rate of cange of V wen r = 3 cm. 36π For te following exercises, te revenue generated by selling x items is given by R(x) = 2x x. 40/53

41 Find te average cange of te revenue function as x canges from x = 10 to x = 20. Find R'(10) and interpret. $50.00 per unit, wic is te instantaneous rate of cange of revenue wen exactly 10 units are sold. Find R'(15) and interpret. Compare R'(15) to R'(10), and explain te difference. For te following exercises, te cost of producing x cellpones is described by te function C(x) = x 2 4x Find te average rate of cange in te total cost as x canges from x = 10 to x = 15. $21 per unit Find te approximate marginal cost, wen 15 cellpones ave been produced, of producing te 16 t cellpone. Find te approximate marginal cost, wen 20 cellpones ave been produced, of producing te 21 st cellpone. $36 Extension For te following exercises, use te definition for te derivative at a point x = a, f(x) f(a) x a, to find te derivative of te functions. lim x a f(x) = 1 x 2 f(x) = 5x 2 x + 4 f'(x) = 10a 1 f(x) = x 2 + 4x + 7 f(x) = 4 3 x 2 4 (3 x) 2 41/53

42 Capter Review Exercises Finding Limits: A Numerical and Grapical Approac For te following exercises, use [link]. lim f(x) x lim f(x) x 1 lim f(x) x 1 does not exist lim f(x) x 3 42/53

43 At wat values of x is te function discontinuous? Wat condition of continuity is violated? Discontinuous at x = 1( lim f(x) does not exist), x = 3 (jump discontinuity), x a and x = 7 ( lim f(x) does not exist). x a Using [link], estimate lim f(x). x 0 3 x F(x) Undefined For te following exercises, wit te use of a graping utility, use numerical or grapical evidence to determine te left- and rigt-and limits of te function given as x approaces a. If te function as limit as x approaces a, state it. If not, discuss wy tere is no limit. f(x) = { x 1, x 3, if if x 1 x = 1 a = 1 f(x) = { 1 x + 1, (x + 1) 2, if if x = 2 x 2 a = 2 lim x 2 f(x) = 1 f(x) = { x + 3, 3 x, if if x < 1 x > 1 a = 1 Finding Limits: Properties of Limits For te following exercises, find te limits if lim f(x) = 3 and lim g(x) = 5. x c x c lim (f(x) + g(x)) x c 43/53

44 2 f(x) lim g(x) x c lim (f(x) g(x)) x c 15 { lim f(x), f(x) = 3x 2 + 2x + 1 x 0 + 5x + 3 x > 0 x < 0 { lim f(x), f(x) = 3x 2 + 2x + 1 x 0 5x + 3 x > 0 x < 0 3 lim (3x x ) x 3 + For te following exercises, evaluate te limits using algebraic tecniques. lim 0( 12 lim x 25( x lim x 1( x 10 2 ( + 6) 36 ) x 5 ) 2 9x x ) lim x x + 1 x 4 lim x 3( x 3 + x ) 44/53

45 1 9 Continuity For te following exercises, use numerical evidence to determine weter te limit exists at x = a. If not, describe te beavior of te grap of te function at x = a. f(x) = 2 x 4 ; a = 4 f(x) = 2 (x 4) 2 ; a = 4 At x = 4, te function as a vertical asymptote. f(x) = x x 2 x 6 ; a = 3 f(x) = 6x2 + 23x x 2 ; a = removable discontinuity at a = 5 2 f(x) = x 3 9 x ; a = 9 For te following exercises, determine were te given function f(x) is continuous. Were it is not continuous, state wic conditions fail, and classify any discontinuities. f(x) = x 2 2x 15 continuous on (, ) f(x) = x2 2x 15 x 5 f(x) = x2 2x x 2 4x + 4 removable discontinuity at x = 2. f(2) is not defined, but limits exist. f(x) = x x 2 12x + 10 f(x) = x2 1 x 2 x 45/53

46 discontinuity at x = 0 and x = 2. Bot f(0) and f(2) are not defined. f(x) = x + 2 x 2 3x 10 f(x) = x + 2 x removable discontinuity at x = 2. f( 2) is not defined. For te following exercises, find te average rate of cange f(x + ) f(x). f(x) = 3x + 2 f(x) = 5 0 f(x) = 1 x + 1 f(x) = ln(x) ln(x + ) ln(x) f(x) = e 2x For te following exercises, find te derivative of te function. f(x) = 4x 6 = 4 f(x) = 5x 2 3x Find te equation of te tangent line to te grap of f(x) at te indicated x value. f(x) = x 3 + 4x; x = 2. y = 8x + 16 For te following exercises, wit te aid of a graping utility, explain wy te function is not differentiable everywere on its domain. Specify te points were te function is not differentiable. 46/53

47 f(x) = x x Given tat te volume of a rigt circular cone is V = 1 3 πr2 and tat a given cone as a fixed eigt of 9 cm and variable radius lengt, find te instantaneous rate of cange of volume wit respect to radius lengt wen te radius is 2 cm. Give an exact answer in terms of π 12π Practice Test For te following exercises, use te grap of f in [link]. f(1) 3 lim f(x) x 1 + lim f(x) x 1 47/53

48 0 lim f(x) x 1 lim f(x) x 2 1 At wat values of x is f discontinuous? Wat property of continuity is violated? For te following exercises, wit te use of a graping utility, use numerical or grapical evidence to determine te left- and rigt-and limits of te function given as x approaces a. If te function as a limit as x approaces a, state it. If not, discuss wy tere is no limit f(x) = { 1 x 3, if x 3 + 1, if x 2 x > 2 a = 2 lim f(x) = 5 x 2 2a and lim f(x) = 9 Tus, te limit of te function as x approaces 2 x 2 + does not exist. { f(x) = x 3 + 1, 3x 2 1, x , if if if x < 1 x = 1 x > 1 a = 1 For te following exercises, evaluate eac limit using algebraic tecniques. 1 lim x 5( x x) 1 50 lim 0( ) 2 lim 0( ) 48/53

49 1 For te following exercises, determine weter or not te given function f is continuous. If it is continuous, sow wy. If it is not continuous, state wic conditions fail. f(x) = x 2 4 f(x) = x3 4x 2 9x + 36 x 3 3x 2 + 2x 6 removable discontinuity at x = 3 For te following exercises, use te definition of a derivative to find te derivative of te given function at x = a. f(x) = x f(x) = 3 x f'(x) = 3 3 2a2 f(x) = 2x 2 + 9x For te grap in [link], determine were te function is continuous/discontinuous and differentiable/not differentiable. 49/53

50 discontinuous at 2,0, not differentiable at 2,0, 2. For te following exercises, wit te aid of a graping utility, explain wy te function is not differentiable everywere on its domain. Specify te points were te function is not differentiable. f(x) = x 2 x + 2 f(x) = ex not differentiable at x = 0 (no limit) For te following exercises, explain te notation in words wen te eigt of a projectile in feet, s, is a function of time t in seconds after launc and is given by te function s(t). s(0) s(2) 50/53

51 te eigt of te projectile at t = 2 seconds s'(2) s(2) s(1) 2 1 te average velocity from t = 1 to t = 2 s(t) = 0 For te following exercises, use tecnology to evaluate te limit. sin(x) lim 3x x tan lim 2 (x) 2x x 0 sin(x)(1 cos(x)) lim x 0 2x 2 0 Evaluate te limit by and. lim f(x), were f(x) = { 4x 7 x 1 x 2 4 x 1 x = 1 At wat value(s) of x is te function below discontinuous? f(x) = { 4x 7 x 1 x 2 4 x = 1 For te following exercises, consider te function wose grap appears in [link]. 51/53

52 Find te average rate of cange of te function from x = 1 to x = 3. 2 Find all values of x at wic f'(x) = 0. x = 1 Find all values of x at wic f'(x) does not exist. Find an equation of te tangent line to te grap of f te indicated point: f(x) = 3x 2 2x 6, x = 2 y = 14x 18 For te following exercises, use te function f(x) = x(1 x) 2 5. Grap te function f(x) = x(1 x) 2 5 by entering f(x) = x ( (1 x) 2 ) 1 5 and ten by entering f(x) = x ( (1 x) 1 5) 2. Explore te beavior of te grap of f(x) around x = 1 by graping te function on te following domains, [0.9, 1.1], [0.99, 1.01], [0.999, 1.001], and [0.9999, ]. Use tis information to determine weter te function appears to be differentiable at x = 1. Te grap is not differentiable at x = 1 (cusp). 52/53

53 For te following exercises, find te derivative of eac of te functions using te definition: lim 0 f(x) = 2x 8 f(x) = 4x 2 7 f ' (x) = 8x f(x + ) f(x) f(x) = x 1 2 x2 f(x) = 1 x + 2 f ' (x) = f(x) = 3 x 1 1 (2 + x) 2 f(x) = x f ' (x) = 3x 2 f(x) = x 2 + x 3 f(x) = x 1 f'(x) = 1 2 x 1 53/53