AB Calculus Unit 4.14 Quiz 4.4-4.6 Unit 4 Review Key Ideas

• Absolute (Global) Extreme Values • Local (Relative) Extreme Values • Finding Extreme Values • Mean Value Theorem • Physical Interpretation • Increasing and Decreasing Functions • First Derivative Test for local Extrema • Concavity • Points of Inflection • Second Derivative Test for Local Extrema • Learning about Functions from Derivatives • Modelling and Optimization • Linear Approximation • Newton’s Method • Differentials • Estimating Change with Differentials • Absolute, Relative, and Percentage Change • Related Rates • Relative Motion

Concepts Optimization Determine, by algebraic manipulation, one equation which has the variable to be optimized equal to an expression written in terms of one other variable. The optimal solution is located where the first derivative equals zero (or the endpoints). Check the candidates (derivatives and endpoints) by evaluating them in the original equation to locate maximum or minimum value. Related Rates Determine what rate is being asked for, what rate is given, and find a relationship (equation) that has both of those variables included. Perform the derivative in terms of the rate (usually time) and now insert the values that indicate the instant the rate is required, and solve.

Unit 4.11 AP Specific Practice Question

1. Let ( ) ( )( )

2

2

3nwnw

+=f w , where n>0. Which of the following statements in not true?

A. y=1 is a horizontal asymptote of f. B. f is decreasing for w>0.

C. f has a critical point at 3wn

= −

D. w=0 is a vertical asymptote of f. E. f has no relative maxima or minima for w<0 2. The height of a cylinder is increasing at a rate of 6 cm per second. The radius of the cylinder is

decreasing at a rate of 2 cm per second. The volume is: A. always increasing. B. always decreasing

C. increasing when 23

<r h

D. decreasing when 23

<r h

E. constant 3. Sketch a function with the following properties: x<0 0<x<1 1<x + - - ( )f x′

+ + - ( )f x′′

4. Determine the point of inflection for the function ( ) 2 xf x xe=

5. The curve below is the graph of on the domain [ ] ( )y f x= 0, 6

]

Which of the following statements are true about f? I. There is some point c in the interval [ such that . 0, 6 ( ) 0f c′ = II. f is continuous at x=3. III. f ′ ( )5 0< A. I only B. II only C. I and II only D. II and III only E. I, II, and III

6. Suppose ( ) ( )322x= +f x . Use tangent line approximation to approximate . ( )2.1f

7. The velocity of an object moving along the x-axis is given by where

t>0. At what time (t>0) does the object’s maximum acceleration occur? ( ) 3 23 2v t t t t= − + − +1

8. A family is at an air show and watches a jet approach. The jet approaches at a rate of 12km per

minute at an altitude of 8 km. a) At what speed (in radians per minute), is the angle of the altitude changing when the

straight-line distance from the family to the jet is 32 km? b) At what rate is the straight-line distance between the family and the jet changing when the

jet is 10 km from the family?

Solutions 1. E.

( ) ( )( )

( )( )

2 2 2

2

2

3 3 31

3 32 1

nw nwf wnw nwnw

f wnw nw

+ + = = = +

′ = + ⋅ −

Therefore f(w) has a relative minimum at 3wn

= −

2 D.

( ) ( )( ) ( ) ( )[ ]

( )

2

22

2 2 24 6

V r hdV dr dhr h rdt dt dt

r h rr h r

π

π π

π π

π

= = +

= − + = − +

6

Therefore 0dV< when

dt23

r h< .

3. Infinite possibilities 4. The point of inflection candidate occurs when ( ) 0f x′′ = ( )

( ) [ ]( ) ( )( ) [ ]( ) ( )

( )( )

22 2

2 2 2

4 22 2

0 2 22

x

x x

x x

x x

x

x

f x xef x e x e

f x e e x ee xee xe x

x

=

′ = + ′′ = + +

= += += += −

x

0

Now to check by picking x-values on either side to ensure concavity changes

. ( ) ( )3 0, 0f f′′ ′′− < > Therefore the point of inflection occurs when . 2x = −

5. D There is no horizontal piece in the graph. 6. ( ) ( )( )

( )( ) ( )y f a f a x ay f a x a f a

′− = −′= − +

( ) ( )

( ) ( )

32

12

23 22

f x x

f x x

= +

′ = +

We notice that when we use 2 we can determine the exact answer. Therefore a=2 and x=1.99 ( )( ) ( )

( ) ( )( ) ( )( )

2 2.1 2 22.1 2 2.1 2 2

3 0.1 88.3

y f ff f f

′= − +′= − +

= +=

7. We want maximum acceleration, therefore we must set the derivative of the acceleration

function to zero and solve for the candidates. ( )

( )( )

3 2

2

3 29 4 118 4

v t t t ta t t ta t t

= − + − += − + −

′ = − +

1

0 18

18 429

tt

t

= − +=

=

4

0 0

We note that when t=0, and when t=1, , therefore ( )0a′ > ( )1a′ <29

a is a maximum.

8a.

θ

8

x

z This is a related rate question

12

?, when z=32

dxdtddtθ

= −

=

( )( )

( )2

2

tan8

8 tan

8sec

328 128

128 12

3 radians per minute32

x

xd dxdt dtddtddtddt

θ

θθ

θ

θ

θ

θ

=

=

⋅ =

=

=

=

8b. We require dz when z=10 dt

2 2 2

2 2 2

z x ydz dx dyz x ydt dt dtdz dx dyz x ydt dt dt

= +

= +

= +

We know y=8, and z=10. therefore 100 64

6x = −

=

( ) ( )10 6 12 8 0

7.2 km per minute

dzdtdzdt

= − +

= −