Some Review Problems for First Midterm Mathematics 1300, Calculus 1

Transcription

1 Some Review Problems for First Midterm Matematics 00, Calculus. Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd, periodic function tat as been sifted upwards, so we will use sin(t) as our parent function and apply te appropriate transformations. Te grap appears ( to be ) a function wose amplitude is, period is π, t and midline is. So f(t) = sin +.. Find te domain of te function f() = 5. Because te square root of a negative number doesn t produce real numbers, we will only allow our domain to be values of suc tat 5 is not negative. In oter words, 5 0. We solve tis inequality for : So te domain is (, /5] Wat is te domain and range of te function f() = ( + )? Tis function is a polynomial, so its domain is all real numbers. Te sape of te grap is a parabola tat opens upwards, so to find te range we first find te verte. Te equation of f() is already in verte form, so we know te verte is (, ). Since te grap as a minimum at its verte and etends upwards infinitely, te range is [, ).. Use te it definition of derivative to compute te derivative of te function f() = at =. Te it definition of a derivative is f f( + ) f() (). Using tis definition,

2 f () f( + ) f() + ( ) ( +) + + [Since f() = ] [Common denominators] (+) (+) (+) (+) (+) (+) (+) ( + ) ( ) [Distributive property] [Dividing by is te same as multiplying by ] ( + ) ( + ) = ( + 0) [Cancellation of ] [Evaluate it by setting = 0] = [Evaluate it by setting = 0] Te derivative at = is f () = ( ) =. 5. Te table below gives te dept of snow tat as fallen in inces as a function of time in ours past 8am. Wat is d(.5) (including units) and wat does it represent? Wat

3 time did it start snowing? Net, estimate d () and d (). Include units in your answer and say in a full Englis sentence wat te meaning of tese numbers are d() d(.5) =.5 inces, wic represents te dept of te snow at.5 ours past 8 am, (wic is 9:0 am). Since d(.75) = 0 inces, and after tat time te dept of snow is positive, we can conclude tat it started snowing at 8 : 5 am. We can estimate d () by calculating te slope between = and =.75: d() d(.75).75 = =. inces/r. We can estimate d () in te same way, by calculating te slope between = and =.75: d() d(.75).75 = = 0.6 inces/r. Te interpretation is tat d (). inces/r means tat at 9 am, te dept of te snow was increasing at. inces per our. Likewise, d () 0.6 inces/our means at 0 am, te dept of te snow was increasing at.06 inces per our. 6. Say f() = +. Use te definition of derivative to find f (). Ten write te equation of te tangent line to te curve at =. f () f( + ) f() ( + ) + ( + ) (() + ) = 7. f () = 7 is te slope of te tangent line at =, and f() =, so te equation of te tangent line to te curve is y = 7( ). 7. Say tat te position of an object moving orizontally is given by s(t) = 6t + 7, were position is measured in miles and time is measured in ours. Find te instantaneous velocity at an arbitrary t = a, and ten te instantaneous velocity at t = 7. Include units.

4 Te instantaneous velocity at a time t is given by s (t), s (a) s(a + ) s(a) s 6 (7) = = 6 6(7)+7 7 6(a + ) + 7 6a + 7 6(a + ) + 7 6a + 7 ( ) 6(a + ) a + 7 6(a + ) a + 7 6(a + ) + 7 (6a + 7) ( 6(a + ) a + 7) 6 ( 6(a + ) a + 7) 6 ( 6(a + ) a + 7) 6 6a + 7 =. [Tis is instantaneous velocity at t = a] 6a + 7 miles/our. [Tis is instantaneous velocity at t = 7] 8. Consider te function + cos wen 5 < < 0 f() = 5 wen = 0 e / wen 0 < < 5 (a) Define wat it means for a function f() to be continuous at a point = a. A function f() is continuous at a point = a wen f() eists a and it is equal to f(a). (b) (c) (d) f() = + f() = 0 + f() = 0 (e) 0 f() = e/ = e. + e/ = e 0 = cos = 0 + cos 0 =. 0 0 f() f() =. + (f) Is f() continuous at = 0? Fully eplain your answer. If it is not continuous, state wat type of discontinuity it is and wy. Te it as approaces zero eists (we calculated it in part (e)), but it is not equal to f(0). Terefore, f() is not continuous at = 0. Note tat te discontinuity at = 0 is a removable discontinuity because te f() 0 eists.

5 9. Sketc te grap of a function g() tat satisfies ALL of te following properties: (a) g() = + (b) g() does not eist. (c) g() = (d) (e) g() = + g() = A car is first driven at an increasing speed and ten te speed begins to decrease until te car stops. Sketc a grap of te distance te car as traveled as a function of time. Assuming tat te car is moving in te forward direction, wen te car s speed is increasing, tis means te velocity increasing. Tat implies tat te second derivative is positive, so te grap of position sould be concave up. Similarly, wente speed is decreasing, te velocity is decreasing. Tat implies tat te second derivative is negative, so te grap of position is concave down. Ten notice tat wen te car stops, position sould become constant. Te grap could look like tis: 5

6 s(t) t. Find te average velocity over te interval 0. t 0. of a car wose position s(t) is given by te following table. Ten estimate te velocity at t = 0.. Include units. t (sec) s(t) (ft) Te average velocity between t = 0. and t = 0. is s(0.) s(0.) = = 5 ft/sec. To estimate te velocity at t = 0. better, let us also calculate te average velocity between t = 0. and t = 0. and ten average te two average velocities. Te average velocity between t = 0. and t = 0. is s(0.) s(0.) =.8 0. = 8 ft/sec. Ten taking te average, = 6.5 ft/sec, so 6.5 feet per second is an estimate for te velocity at t = 0... Te function f(t) = 6t + 6t gives te distance above te ground of a ball tat is trown from ground level straigt up into te air at time t = 0, wit an initial velocity of 6 ft/sec. (a) How ig is te ball above te ground at t = second? f() = 6() + 6() = 8 ft. (b) How fast is te ball moving at t =, t =, t = and t = seconds? First we calculate te derivative, f (t) by using te it definition. After we do tis, we sould get tat f (t) t + 6. To calculate te velocity at 6

7 t =, t =, t =, and t =, we substitute tese values into f (t). f () = + 6 = ft/sec f () = () + 6 = 0 ft/sec f () = () + 6 = ft/sec f () = () + 6 = 6 ft/sec. (c) Wat is te maimum eigt of te ball? (Hint: you sould ave gotten a velocity of 0 for one of te values of t in te last part) Since te ball travels upward and ten starts to fall, we see tat te maimum occurs wen te ball as zero velocity at te peak of its trajectory. At time t =, te velocity is zero, so tis must be te point at wic te ball stops traveling upward (i.e. positive velocity) and begins its traveling downward (i.e. negative velocity). Te maimum eigt is te eigt at t =, so f() = 6() + 6() = 6 ft.. Precisely state te Intermediate Value Teorem. A polynomial p() as p( ) = and p(0) = and p() =. Sow tat p() as at least two zeroes. IVT: If f() is continuous on a closed interval [a, b] and N is any number between f(a) and f(b) (were f(a) f(b)), ten tere is a number c in (a, b) suc tat f(c) = N. Since p() is polynomial, it is continuous everywere. Tus we may use te IVT. p( ) = and p(0) =. Since 0 is between and -, te IVT states tat tere must eist a value c between - and 0 suc tat p(c ) = 0. So p() as a zero between and 0. Similarly, we know p(0) = and p() =. Again, 0 is between - and, so tere must eist a value c between 0 and suc tat p(c ) = 0. So p() as a zero between 0 and. Tus p() as at least two zeros.. Find te following its. Sow all of your work. + (a) + ( )( + 6) ( ) ( + 6) = 8 7

8 (b) ( ) ( + ) ( 9) ( + ) 9 ( 9)( + ) + = 6 (c) = 8 ( ) () ( ) () ( )() + (d) + ( + ) ( ) + = = ( ) ( ) (e) + 8

9 + ( + ) ( ) ( ) ( ) + = + (f) = = + ( ) ( + ) ( ) (g) ( 9 + ) ( 9 + ) = + = (Warning: Wile it is true tat + =, it is not true tat = 0, Observe te following problem.) () ( 9 + ) 9

10 ( ( 9 + ) 9 + ) (9 + ) = ( ( ) ) ( ) ( ) ( ) = 6 5. Te curve below sows position s, measured in feet, as a function of time t, measured in seconds. Te dotted line is tangent to te curve. 6 s(t) 5 t (a) (b) Find te average velocity over te interval from t = 0 to t =. Include units. s() s(0) average velocity = = 0 5 = 5 0 ft/s Find te instantaneous velocity at t =. Include units. Te instantaneous velocity is te slope of te tangent line at t =. Tis line goes troug (, 5) and (, ). So te instantaneous velocity is s () = 5 = ft/sec. 0

11 (c) Find te equation of te tangent line. Te equation of a line troug (, ) wit slope - is s = (t ) or s = t Here s a grap of te function f(): 5 Meanwile, ere is a definition of te function g(): ( ) + > g() = + 6 < (a) Eplain wy te grap defines f as a function of. It passes te vertical line test. (b) Is te f function invertible? No, it fails te orizontal line test. (c) Is te function g invertible? By sketcing g, we can see tat it is inveritble. 5 (d) Wat is te domain and te range of f? Te domain of f is (, ) (, ) (, 5) and te range of f is (, ). (e) Wat is te domain and te range of g? Te domain of g is (, ) (, ) and te range of g is (, ) (, ).

12 (f) Find te it of f() as approaces -, as approaces, as approaces, as approaces, as approaces. Clearly eplain your results. f() = DNE; f() = but + f() = ; removable discontinuity f() = ; f is continuous at f() = DNE; f() = but + f() = ; removable discontinuity f() = f() = (g) Wat is te it of g() as approaces? Wat about as approaces 0? 0 g() = DNE; g() = but + g() = 6; g is continuous at 0 g() = () List all of te discontinuities of f and g and clearly eplain wy tese are discontinuities. State wat type of discontinuities tey are and wy. Discontinuities of f: = : f() eists, but f( ) is undefined, a removable discontinuity = : f() =, an infinite discontinuity = : f() = but f() =, a removable discontinuity = : f() f(), a jump discontinuity since bot sided its eist + = : f() = but f() is undefined, a removable discontinuity = 5 : f() eists, but f(5) is undefined, a removable discontinuity 5 Discontinuities of g: = : g() g(), a jump discontinuity since bot sided its eist +

13 (i) (f() + g()) = (f() + g()) f() + g() = + = 5 (j) (f() + g()) = +(f() + g()) f() + g() = + = (f() + g()) f() + g() = + = 5 (f() + g()) = 5 7. Find two values of te constant b so tat te following function is continuous: + b () = + < b For () to be continuous, we need () (). b + b () = b + b and () = b +. Setting tem equal gives, b + b b + b = b + b + b = 0 (b + 6)(b ) = 0 b = 6 or b =. We ave 8. If 9 f() + 7 for 0, ten find f(). By te Squeeze Teorem, Tus, f() = 7. 9 f() + 7 () 9 f() () f() Suppose f() = and f () =. Find f( ) and f ( ) if f is assumed to be even. f( ) = and f ( ) =.

14 0. Given all of te following information about a function f, sketc its grap. f() = 0 at = 5, = 0, and = 5 f() = f() = f () = 0 at =, =.5, and = 7. Given te grap of y = f() sown, sketc te grap of te derivative Given te following table of values, estimate f (0.6) and f (0.5); ten use tese to estimate f (0.6) f() f (0.6).5 and f (0.5), so f (0.6) 7.5.

15 . If possible, draw te grap of a differentiable function wit domain [0, 6] satisfying f () > 0 for <, f () < 0 for >, f () > 0 for >, f () > 0 for <, and f () < 0 for >. It s impossible. If f() is differentiable at =, ten te fact tat f () > 0 for < means tat f () is increasing, so f () can t go from positive on [0, ) to negative on (, ).. If y = f() is te grap sown below, sketc te graps of y = f () and y = f (). Estimate slopes as needed y = f () y = f () Te grap of y = f () is sown. Were is f() greatest? Least? Were is f () greatest? Least? Were is f () greatest? Least? Were is f() concave up? Concave down? 5

16 0 0 f () is positive for te entire interval, so f() is greatest at.5 and least at.5. Looking at te peaks and valleys of te grap of f () given, f () is greatest at = 0. and least at =.. Looking at te slope of f (), we estimate f () is greatest at = and =.5, and f () is least at =.5. f() is concave up were f () is increasing, from about. to 0. and about.9 to.5. f() is concave down were f () is decreasing, from about.5 to. and from about 0. to Name tree different reasons a function can fail to be differentiable at a point. It could be discontinuous, or ave a vertical tangent like y = /, or ave a corner like y =. 6