1. The cost (in dollars) of producing x units of a certain commodity is C(x) = x x 2.

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1 APPM 1350 Review #2 Summer The cost (in dollars) of producing units of a certain commodity is C() (a) Find the average rate of change of C with respect to when the production level is changed i. from 100 to 105 ii. from 100 to 101 (b) Find the instantaneous rate of change of C with respect to when 100. C(105) C(100) (a) (i) Average rate of change C(101) C(100) (ii) Average rate of change (b) The instantaneous rate of change is C (100) where C () , so C(100) 20 { 2. Determine whether f (0) eists given f() 2 sin(1/), if 0 0, if 0 We first need to check if f() is continuous at 0, so note that 1 sin(1/) 1 and so 2 2 sin(1/) 2 and since lim 2 lim 2 0, we have that lim 2 sin(1/) 0 by the Squeeze Theorem, that is, lim 2 sin(1/) f(0) and so f() is continuous at 0. 0 Now we need to show that f (0) eists, i.e. that the derivative from the left and right of 0 eist and are equal. Note that from the left of 0, we have f(0 + h) f(0) h 2 sin(1/h) 0 lim lim lim h 0 h h 0 h h sin(1/h) h 0 and recall that 1 sin(1/h) 1 and so if h < 0 then h h sin(1/h) h and now since lim h lim h 0 then lim h sin(1/h) 0 by the Squeeze Theorem. Similarly, we can show h 0 h 0 h 0 lim h sin(1/h) 0 by the Squeeze Theorem and thus f (0) 0. h Find an equation of the tangent line to y 2 2 at the point (0, 0). + 1 Note that f () 2 22 ( 2 + 1) 2 and f(0) 0 and f (0) 2 so the equation of the tangent line is y f(0) + f (0)( 0) 2, i.e. y If a particle has position function s(t) t 3 9t t + 10, where t is measured in seconds and s in feet. Find the total distance traveled in the first 8 seconds. Note that v(t) s (t) 3t 2 18t (t 2 6t + 5) 3(t 1)(t 5), so v(t) 0 when t 1 and t 5, so the total distance traveled in the first 8 seconds is Total Distance s(8) s(5) + s(5) s(1) + s(1) s(0) 120ft 5. If f is a differentiable function, find the derivative of y 1 + f(). f () (f() + f ()) 1 2 1/2 (1 + f()) 22 f () + f() 1 2 3/2

2 6. Find the first and second derivatives of the function: (a) F (t) (1 7t) 6 (b) y ( 3 + 1) 2/3 (c) y (a) F (t) 42(1 7t) 5 and F (t) 1470(1 7t) 4 (b) y 2 2 ( 3 + 1) 1/3 and y 2(3 + 2) ( 3 + 1) 4/3 2 ( 1) 2 (c) y 2( + 1) ( 1) 3 and y 4( + 2) ( 1) 4 7. Find dy/d: (a) y 5 (b) y sin( 2 ) sin(y 2 ) (c) y 1+ 2 y (d) y y y (a) y 5 2 ( 5) 3/2 (b) y sin(y2 ) 2y cos( 2 ) sin( 2 ) 2y cos(y 2 ) (c) y y 4(y)3/2 2 2 y (d) y 43 y 2y 3 5y y A man starts walking north at 4 ft/s from a point P. Five minutes later, a woman starts walking south at 5 ft/s from a point 500 ft due east of point P. At what rate are the people moving apart 15 minutes after the woman starts walking? h h P 500 y P y Figure 1: Man and womans path 500 Figure 2: Man and womans path organized as a triangle Let: the distance the man has walked, then d 4 ft/s y the distance the woman has walked, then dy 5 ft/s h the distance between the man and the woman Here we have h 2 ( + y) , and we wish to find dh when t 15 min 900 sec. Recall that distance rate time and so when t 900 sec, we have 4( ) 4800 ft }{{} 5 min headstart y 5(900) 4500 ft h 2 ( + y) ( ), so h

3 And note, h 2 ( + y) implies 2h dh ( ) d + dy dh 2( + y) 2h ( d 2( + y) + dy ) and so 9300(4 + 5) (9) ft/sec 9. Water is leaking out of an inverted conical tank at a rate of 10,000 cm 3 /min at the same time that water is being pumped into the tank at a constant rate. The tank has height 600 cm and the diameter of the top is 400 cm. If the water level is rising at a rate of 20 cm/min when the height of the water is 200 cm, find the rate at which water is being pumped into the tank. Let: K rate at which water is pumped in (cm 3 /min) r radius of water h height of water dh 20 cm/min when h 200 cm leak 10, 000 cm 3 /min We wish to find K, note that V 1 3 πr2 h and since we have similar triangles we have r h r h 3 so V 1 27 πh3 Now note that dv rate in rate out K 10000, also note that V 1 27 πh3 implies dv 1 dh πh2 9 And so K 10, dh πh2 and thus 9 K 1 dh πh , 000 π 800, 000π 9 (200)2 (20) + 10, , 000 cm 3 /min Find the linearizaton L() of f() at 0 and use it to approimate Note that L() f(0) + f (0) 1 + 1, and f( 0.05) L( 0.05) L( 20 ) / Find the absolute maimum and minimum values of f on the given interval: (a) f() , [ 1, 4] (b) f() sin() + cos(), [ π/4, π/2] (a) Note that f () ( 1)( 3) so the critcal points are 1 and 3 and f(1) 6, f(3) 2 and f( 1) 14, f(4) 6. So, the absolute maimum value is 6 at 1 and 4 and the absolute minimum value is -14 at 1. (b) Here f () cos() sin() and f () 0 implies cos() sin() which implies tan() 1 so π/4 is a critical point, and f( π/4) 0, f(π/4) 2 and f(π/2) 1, so the absolute maimum value is 2 at π/4 and the absolute minimum value is 0 at π/4.

4 12. State Rolle s Theorem. Now, let f() 1 2/3. Show that f( 1) f(1) but there is no number c in ( 1, 1) which satisfies Rolle s Theorem. Does this contradict Rolle s Theorem? Why or why not? Rolle s Theorem: If f() is continuous on [a, b] and differentiable on (a, b) and if f(a) f(b) then there eist a c in (a, b) such that f (c) 0. Note that here clearly f( 1) 0 f(1) and f () 2 and clearly there is no number c for 31/3 which f (c) 0. This does not contradict Rolle s Theorem since f() is not differentiable on ( 1, 1). 13. Verify that f() 3 satisfies the hypothesis of the Mean Value Theorem on the interval [0, 1]. Find all values of c that satisfy the conclusion of the Mean Value Theorem. Note that f() 3 is continuous on [0, 1] and f () 1 and so f() is differentiable on (0, 1). 32/3 f(1) f(0) 1 Now note that 1 and f (c) 1 implies c At 2:00 p.m., a car s speedometer reads 30 mi/h. At 2:10 p.m., it reads 50 mi/h. Assuming the car has been driven the whole time, show that at some time between 2:00 and 2:10, the acceleration is 120 mi/h 2. Let v(t) be the speed of the vehicle as a function of time (in hours). Then v(0) 30 and v(1/6) 50, and so by the Mean Value Theorem there eists a time t 0 between t 0 (2 o clock) and t 1/6 (2:10 p.m.) such that a(t 0 ) v (t 0 ) v(1/6) v(0) 1/ /6 120 mi/h Show that the equation 2 1 sin() 0 has eactly one real root. Eistence: Let f() 2 1 sin(), note that f(0) 1 < 0 and f(π/2) π 2 > 0. Since f() is the combination of a polynomial and the sine function, f() is continuous. Now, by the Intermediate Value Theorem, there eists at least one point c in the interval (0, π/2) such that f(c) 0, that is, there eists at least one root of f(). Uniqueness: By way of contradiction, assume there is more than one root, assume there are two roots a and b of f() such that a b. Without loss of generality, we can assume a < b, now we apply the Mean Value Theorem on f() on the interval [a, b]. Note that that f() is continuous on [a, b] and f () 2 cos() which is defined on (a, b) and so f() is differentiable on (a, b), thus by the Mean Value theorem there is a number z between a and b such that f (z) f(b) f(a) b a Now note that 1 cos() 1 implies 1 cos() 1, and so 1 2 cos() 3, that is, f () 2 cos() 1 which contradicts f (z) 0. This contradiction shows that the given equation can t have two distinct roots and so it has eactly one real root. 0

5 16. For the functions given below (a) Find the intervals on which f is increasing or decreasing (b) Find the local maimum and minimum values of f (c) Find the inflection points and state where the function is concave up/down. (i) f() (ii) f() 2 2 (iii) f() 2 sin(), 0 2π + 3 (i) Note that f () ( 3 1) 4( 1)( ) and f () 12 2 so (a) f() decreases on (, 1) and increases on (1, ) (Note, the only real root of f () is 1) (b) There is a local minimum at 1 with value f(1) 4 (Note 1 is in the domain of f()) (c) f() is concave up on (, 0) and (0, ), there is no inflection point. First derivative test for (i) 1 Concavity test for (i) 0 (ii) Note that f () 6 ( 2 + 3) 2 and f () 18(2 1) ( 2 + 3) 3 (a) f() decreases on (, 0) and increases on (0, ) (b) There is a local minimum at 0 with value f(0) 0. (c) f() is concave up on ( 1, 1) and concave down on (, 1) and (1, ). There are inflections points at ( 1, 1/4) and (1, 1/4). First derivative test for (ii) 0 Concavity test for (ii) -1 1 (iii) Note that f () 1 2 cos() and f () 2 sin() (a) f() decreases on (0, π/3) (5π/3, 2π) and increases on (π/3, 5π/3). (b) There is a local minimum at π/3 with value f(π/3) π/3 3. There is a local maimum at 5π/3 with value f(5π/3) 5π/3 + 3 (c) f() is concave up on (0, π) and concave down on (π, 2π). There is an inflection point at (π, π). First derivative test for (iii) 0 π/3 5π/3 2π Concavity test for (iii) 0 π 2π/

6 17. For the functions given below, state the domain, find all horizontal and vertical asymptotes, find all local maima, minima, and inflection points and then sketch the graph (a) y ( 1) 2 (b) y (c) y 5 (a) Note that f () 2( + 1) ( 1) 3 and f 4( + 2) () ( 1) Domain: 1, Horizontal Asymptote: y 0, Vertical Asymptote: 1 Local Maimum Point: None, Inflection Point at ( 2, 4/9) (b) Note here f () Local Minimum point at (, y) ( 1, 1/2), and f () (32 2) 2 3 (1 2 ) 3/ Domain: [ 1, 0) (0, 1], Horizontal Asymptote: None, Vertical Asymptote: 0 Local Maimum point: None, Local Minimum point: None,

7 ( ) ( ) Inflection Points at 3, and 2 3, 2 (c) Note here f () 5 2 ( 5) 3 and f () 5(4 5) 4 3 ( 5) Domain: (, 0] (5, ), Horizontal Asymptote: y 1, Vertical Asymptote: 5, Local Maimum point: None, Local Minimum point: None, Inflection Points: None