MAT 1339-S14 Class 4

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1 MAT 9-S4 Class 4 July 4, 204 Contents Curve Sketching. Concavity and the Second Derivative Test Simple Rational Functions Putting It All Together Optimization Problems Derivative of Sinusoidal Functions 4. Instantaneous Rate of Change of Sinusoidal Functions Curve Sketching. Concavity and the Second Derivative Test Definition... The graph of f(x) is concave up on the interval a < x < b if all the tangents on the interval are below the curve. The graph curves upward. 2. The graph of f(x) is concave down on the interval a < x < b if all the tangents on the interval are above the curve. The graph curves downward.. A point at which the graph changes from being concave up to concave down, or vice versa, is called an inflection point. The concavity is closely related to the second derivatives as shown in the following theorem. Theorem.2. (The Second Derivative Test). A function is concave up on an interval if f (x) > 0 on that interval. If f (a) = 0 and f (a) > 0, then (a, f(a)) is a local minimum point. 2. A function is concave down on an interval if f (x) < 0 on that interval. If f (a) = 0 and f (a) < 0, then (a, f(a)) is a local maximum point.

2 . If f (a) = 0 and f (x) changes sign at x = a, then (a, f(a)) is an inflection point. The proof of this theorem relies on the geometric meaning of the second derivative: rate of change of the slope of the tangent line. Example.. Let f(x) = 2x x 2 2x + 5. Find the inflection points and the intervals of concavity. Solution: It is easy to see that f (x) = 6x 2 6x 2 and f (x) = 2x 6. Let f (x) = 0, we get x = 2. We can calculate the following table. intervals x < /2 x > 2 2 test value 0 2 f (x) = 2x f(x) inflection point (, ) 2 2 Thus f(x) is concave down on the interval < x < 2 < x < +. The inflection point is (, ) and concave up on the interval Example.4. Use the second derivative test to find the local maximum points and local minimum points of the function f(x) = 2x x 2 2x + 5. Solution 2: We can calculate the following table: x = x = 2 f (x) = 6(x 2)(x + ) 0 0 f (x) = 2x 6 + f(x) local max local min Thus (, 2) is the local maximum point and (2, 5) is the local minimum point..4 Simple Rational Functions Here we consider Simple Rational functions (quotient of two polynomials) like, x,... (x )(x ) (x )(x ) 2 x+2, Vertical Asymptotes Definition.5. The line x = a is a vertical asymptote of y = f(x) if 2

3 . lim x a f(x) = ± or 2. lim x a + f(x) = ± or. lim x a f(x) = ±. Example.6. Let f(x) = 2 x+2. For a rational function (quotient of two polynomials), vertical asymptotes occur at x- values for which the function is not defined (True Root of Denominator). The function f(x) is not defined when the denominator x + 2 = 0, namely x = 4. Claim: x = 4 is the vertical asymptote of f(x). This is because lim f(x) = +, x ( 4) + lim f(x) =, x ( 4) why? why? Example.7. Find the vertical asymptotics for f(x) = (x )(x ) and f(x) = x (x )(x )

4 Horizontal Asymptotes Definition.8. The line y = c is a horizontal asymptote of y = f(x) if. lim x f(x) = c or 2. lim x f(x) = c. Example.9. Let f(x) = 2x x 2 2x + 5. Then f(x) has no horizontal asymptote because lim x ± f(x) = ± a finite number. In general, a polynomial function has no horizontal asymptote. Why? Example.0. Find the horizontal asymptote for the following functions (a) x2 2x 8 (b) 2x 8 x 2 (c) 2x 8 x For part (b) we do not have horizontal asymptote but we can notice a new asymptote called: 4

5 Slant Asymptotes Definition.. The line y = mx + b is a slant (oblique) asymptote of y = f(x) if. lim x [f(x) (mx + b)] = 0 or 2. lim x [f(x) (mx + b)] = 0. Example.2. Let f(x) = x2. Then f(x) has a slant asymptote y = x because x lim [f(x) x] = lim x ± x ± [x2 x] = x In general we can find m and b by letting b = m = lim [x x ± x x] = f(x) lim x ± x, lim [f(x) mx]. x ± lim x ± x = 0. Example.. Use the first and the second derivative to analytic the graph of the function f(x) = x2 x given in example.2 5

6 .5 Putting It All Together The purpose of this section is to sketch the curve of polynomial functions and rational functions by using the first and the second derivatives. Example.4. Let f(x) = x x 2 5x + 5. (a) Find the critical points of f(x). (b) Find the intervals where f(x) is increasing, decreasing. (c) Find the local maximum and local minimum points of f(x). (d) Find the intervals where f(x) is concave up, concave down. (e) Find the inflection points of f(x). (f) Does f(x) have asymptotes? Why? (g) Sketch the graph of f(x). Solution (a): The domain of the function is all real numbers. f (x) = x 2 2x 5 = (x 5)(x + ) = 0, we get x = 5 or x =. Since f (x) is defined everywhere, the only critical points are ( 5, f ( )) 5 = ( 5, 40 ) 27 and (, f( )) = (, 8). Solution (b): f(x) is increasing on (, ) ( 5, ) and f(x) is decreasing on (, 5 ) as shown in the following table. intervals x < x = < x < 5 x = 5/ x > 5 5 test value f (x) = (x 5)(x + ) f(x) Solution (c): From the above table we see that (, 8) is the local maximum point and ( 5, 40 ) is the local minimum point. 27 Solution (d): Let f (x) = 6x 2 = 0, we get x =. From the following table we see that f(x) is concave up on (, ) and f(x) is concave down on (, ). intervals x < / x > test value 0 f (x) = 6x f(x) inflection point (, f()) 6

7 Solution (e): From the above table we see (, f( )) = (, 88 ) is the inflection point 27 of f(x). Solution (f): f(x) has no asymptote. () f(x) has no vertical asymptote since lim x a f(x) = a a 2 5a + 5 is a finite number, lim x a f(x) ±. (2) f(x) has no horizontal asymptote because lim x ± f(x) = ± = a finite number. f(x) () f(x) has no slant asymptote because m = lim x ± = lim x x ± x 2 x 5 5 = ± = a finite number. x Solution (g): Example.5. Let f(x) = x x 5. (a) Find the critical points of f(x). (b) Find the intervals where f(x) is increasing, decreasing. (c) Find the local maximum and local minimum points of f(x). (d) Find the intervals where f(x) is concave up, concave down. (e) Find the inflection points of f(x). (f) Does f(x) have asymptotes? Why? (g) Sketch the graph of f(x). Solution (a): The domain of the function is x 5 0, i.e., x 5. Domain: (, 5) (5, ) f (x) = (x) (x 5) x(x 5) (x 5) 2 = x 5 x (x 5) 2 = 5 (x 5) 2. 7

8 So f (x) exists everywhere in the domain of f(x) and f (x) 0 for all x in the domain. So there is no critical points of f(x). Critical points: All x such that f (x) = 0 or f (x) D.N.E but first x should be in the domain first Solution (b): f(x) is decreasing on (, 5) (5, ) as shown in the following table. intervals x < 5 x > 5 test value 0 0 f (x) = 5 (x 5) 2 f(x) Solution (c): Since there is no critical points, there is no local maximum point or local minimum point. Solution (d): Since f (x) = ( 5)(x 2) 2, thus f (x) = ( 5)( 2)(x 5) () = 0 (x 5) (note here we use chain rule, but (x 5) = ). From the following table we see that f(x) is concave up on (5, ) and f(x) is concave down on (, 5). intervals x < 5 x > 5 test value 0 0 f (x) = 0 (x 5) + f(x) There is an inflection on the graph of f around x = 5 but there is no inflection point because x = 5 is not in the domain Solution (e): Since f (x) 0 for all x in the domain of f(x), there is no inflection point of f(x). Solution (f): f(x) has asymptotes. () f(x) has vertical asymptote x = 5 since lim x 5 f(x) = ±. (2) f(x) has horizontal asymptote y = because lim x ± f(x) =. () The slant asymptote of f(x) is the same as the horizontal asymptote because m = f(x) lim x ± = lim x x ± = 0. x 5 8

9 Solution (g):.6 Optimization Problems The purpose of this section is to learn how to solve applied optimization problems. Example.6. A rectangular field is to be enclosed by a fence on sides and by a straight stream on the fourth side. Find the dimensions of the field with maximal area that can be enclosed with 000 feet of fencing. Solution: To translate the problem into mathematical language we need: Step : Introduce variables (a picture may help organize information). length l width w Step 2: Distinguish conditions (identities involving variables) and functions (related to the purpose of the question). function: A=lw condition: 2w+l=000 So the question translates to find the maximal value of the area A. Step : Use condition to simplify function into one variable. 2w + l = 000 = l = 000 2w A = lw = (000 2w)w A(w) = 000w 2w 2 Step 4: Find the critical points of the function. A (w) = 000 4w = 0 = w =

10 Step 5: Find the (absolute) maximum point of the function by st derivative or 2nd derivative test. So the point (250, A(250)) = (250, 25000) is the absolute maximum point of A(w). Conclusion: When w = 250 and l = 000 2w = 500, we get the maximal area of the field Summary: How to solve applied optimization problems? - Introduce variables. 2- Distinguish conditions and functions. - Use condition to simplify function into one variable. 4- Find the critical points of the function. 5- Find the (absolute) maximum (or minimum) point of the function. (st or 2nd derivative test) Example.7. A soup can of volume 500 cm is to be constructed. The material for the top costs 0.4 cent/cm 2. while the material for the bottom and sides costs 0.2 cent/cm 2 Find the dimensions that will minimize the cost of producing the can. 0

11 4 Derivative of Sinusoidal Functions 4. Instantaneous Rate of Change of Sinusoidal Functions In this section, we will review some background of the sine and cosine functions. The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. As shown in the following diagram we have sin x = a c, cos x = b c. The sine and cosine functions give important examples of periodic functions, i.e., functions that repeat the values in regular intervals or periods. We need to be familiar with the graphs of the sine and cosine functions and their variations.

12 It is important to be familiar with the values of the sine and cosine functions at special angles such as x = π, π, π, π. For this we introduce the useful unit circle //204 upload.wikimedia.org/wikipedia/commons/c/c9/unit_circle_angles.svg / There are many important identities involving the sine and cosine functions. For example: sin 2 (x) + cos 2 (x) = ( ) ( ) π π sin 2 x = cos x, cos 2 x = sin x sin(x + h) = sin x cos h + cos x sin h sin(2x) = 2 sin x cos x cos 2x = cos 2 (x) sin 2 (x) 2