ab calculus – step-by-step solution version 1. limits · © 2012 - 2 - illegal to post this...

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AB Calculus – Step-by-Step Solution Version 1. Limits

Consider the functions

�

f x( ) = ax2 − bx2 − 4

and g x( ) = x2 − cx + dx2 − 4

.

a. If

�

limx→ 3

f x( ) = 6 , find a relationship between a and b. (1)

limx→3

f x( ) = 9a − b9 − 4

= 6 ⇒ 9a − b = 30 or b = 9a − 30 1 pt for relationship

b. If, in addition,

�

limx→1

f x( ) = 1, find the values of a and b. (3)

limx→−1

f x( ) = a − b1− 4

= 1⇒ a − b = −3 or b = a + 3

a + 3= 9a − 30 ⇒ 8a = 33⇒ a = 338

b = a + 3= 578

1 pt for b = a + 31 pt for a1 pt for b

c. Find

�

limx→−∞

f x( ) . (1)

limx→−∞

f x( ) = a = 338 1 pt for answer

d. If

�

limx→ 2

g x( ) = 3, find c and d. (4)

limx→2

g x( ) = x2 − cx + dx − 2( ) x + 2( ) = 3

For this to happen, x2 − cx + d must factor into x − 2( ) x − k( )

limx→2

g x( ) = x − 2( ) x − k( )x − 2( ) x + 2( ) =

x − kx + 2

= 2 − k4

= 3

2 − k = 12 ⇒ k = −10x2 − cx + d = x − 2( ) x +10( ) = x2 + 8x − 20c = −8,d = −20

1 pt for realizing numerator must factor into factor into x − 2( ) x − k( )1 pt for x2 + 8x − 201 pt for c, 1 pt for d


AB Calculus – Step-by-Step Solution Version 2. Average/Instantaneous Rate of Change

The Newton Bridge is a toll bridge that gets a lot of traffic during the two hour rush hour period starting at 4:00 PM. The total number of vehicles that go through the northbound is a differentiable function V of time t. A table of selected values of V is given for the time interval

�

0 ≤ t ≤ 2 where t = 0 corresponds to 4 PM.

t (hours) 0 0.5 1 1.5 2

�

V t( ) (vehicles) 0 200 425 755 unknown The southbound tollbooth also monitors vehicles that pass through it by estimating the rate of vehicles that come through. This rate is given by the differentiable function R of t. A table of selected values of R is given for the time interval

�

0 ≤ t ≤ 2 where t = 0 corresponds to 4 PM.

t (hours) 0 0.5 1 1.5 2

�

R t( ) (vehicles per hour) 390 425 425 535 460

a. Approximate

�

′ V 0.75( ). Show the computation that leads to your answer. Explain the meaning of your answer in context to the problem situation using correct units of measure. (3)

′V 0.75( ) ≈ V 1( )−V 0.5( )1− 0.5

= 425 − 2000.5

= 450

Vehicles are passing through the northbound tollbooth at 4:45 PM at approximately 450 vehicles/hour.

1 pt for computation1 pt answer1 pt units

b. Approximate

�

′ R 1.75( ) . Show the computation that leads to your answer. Explain the meaning of your answer in context to the problem situation using correct units of measure. (3)

′R 1.75( ) ≈ R 2( )− R 1.5( )2 −1.5

= 460 − 5350.5

= −150

The rate of vehicles that are passing through the southbound tollbooth at5:45 PM is decreasing at approximately 150 vehicles per hour per hour.

1 pt for computation1 pt answer1 pt units

c. The number of vehicles that have passed through the northbound tollbooth by 6 PM is unknown. It is known that the average rate of change of vehicles passing through this booth between 4 PM and 6 PM is the same as the average rate of change of vehicles passing through this booth between 5 PM and 6 PM. Approximate the total vehicles having passed through the northbound tollbooth by 6:00 PM. Show the computation that leads to your answer. (2)

V 2( )−V 0( )

2 − 0=V 2( )−V 1( )

2 −1⇒V 2( )− 0

2=V 2( )− 425

12V 2( )− 850 =V 2( )⇒V 2( ) = 850

1 pt for computation1 pt answer

d. (1)

�

1 pt for context for parts a and b 4 : 45 PM, 5 : 45 PM( )


AB Calculus – Step-by-Step Solution Version 3. Basic Derivatives and Limits

For the following problems,

�

f x( ) = 2x2 and g x( ) = x2 − 6 .

a. Find

�

limx→−∞

f x( )g x( ) . (1)

limx→−∞

f x( )g x( ) = limx→−∞

2 x2 − 6( )x2

= 2 1 pt answer

b. Find

�

ddx

f x( )g x( )[ ]. (1)

ddx

2x2 −12x2

⎡

⎣⎢

⎤

⎦⎥ =

ddx

2 − 12x2

⎛⎝⎜

⎞⎠⎟ =

12 2x( )x4

= 24x3 1 pt answer

c. Find

�

ddx

x ⋅ g f x( )( )[ ] . (2)

x ⋅g f x( )( ) = x 2x2

⎛⎝⎜

⎞⎠⎟2

− 6⎡

⎣⎢

⎤

⎦⎥ =

4x3

− 6x

ddx

4x3

− 6x⎛⎝⎜

⎞⎠⎟ =

−4 3x2( )x6

− 6 = −12x4

− 6

1 pt for finding x ⋅g f x( )( )1 pt answer

d. If

�

1y

= f x( ) + 1, find dydx

. (2)

y = 1f x( ) +1 =

12x2

+1= x2

x2 + 2

dydx

=x2 + 2( ) 2x( )− x2 2x( )

x2 + 2( )2= 4x

x2 + 2( )2

1 pt for y = x2

x2 + 2

1 pt for dydx

e. Find

�

limΔx→0

′ f −2 + Δx( ) − ′ f −2( )Δx

. (3)

This is asking for the derivative of ′f x( ) or ′′f x( ) at x = −2

′f x( ) = −4x3 so ′′f x( ) = 4 3x2( )

x6 = 12x4

′′f −2( ) = 1216

= 34

1 pt for realizing is ′′f x( ) is needed1 pt for ′′f x( )1 pt for ′′f −2( )


AB Calculus – Step-by-Step Solution Version 4. Basic Derivatives

x

�

f x( )

�

′ f x( )

�

g x( )

�

′ g x( ) 1 4 -3 5 2 2 -3 -1 4 6 3 π 8 -1 4 4 -5 Unknown 0 3

The functions f and g are differentiable for all real numbers g. The table above gives values of the function and their first derivatives at selected values of x.

a. If the function h is given by

�

h x( ) =f x( )g x( )

+ x , find

�

′ h 1( ) . (2)

′h x( ) = g x( ) ′f x( )− f x( ) ′g x( )g x( )⎡⎣ ⎤⎦

2 +1

′h 1( ) = g 1( ) ′f 1( )− f 1( ) ′g 1( )g 1( )⎡⎣ ⎤⎦

2 +1= 5 −3( )− 4 2( )52

+1= −2325

+1= 225

1 pt for ′h x( )1 pt for answer

b. If the function r is given by

�

r x( ) = −2 f x( )g x( ), find the equation of the tangent line to

�

r x( ) at x = 2. (3)

r 2( ) = −2 f 2( )g 2( ) = −2 −3( ) 4( ) = 24

′r x( ) = −2 f x( ) ′g x( ) + g x( ) ′f x( )⎡⎣ ⎤⎦′r 2( ) = −2 f 2( ) ′g 2( ) + g 2( ) ′f 2( )⎡⎣ ⎤⎦ = −2 −3( ) 6( ) + 4 −1( )⎡⎣ ⎤⎦ = 44

y − 24 = 44 x − 2( ) or y = 44x − 64

1 pt for r 2( )1 pt for ′r 2( )1 pt for tangent line equation

c. If the function v is given by

�

v x( ) =f x( ) −1f x( )

, find the slope of the line normal to v at x = 3. (2)

v x( ) = f x( )−1f x( ) = 1− 1

f x( )

′v x( ) = f x( ) 0( )− −1( ) ′f x( )f x( )⎡⎣ ⎤⎦

2 ⇒ ′v 3( ) = ′f 3( )f 3( )⎡⎣ ⎤⎦

2 = 8π2

slope of normal line is −π2

8

1 pt for ′v 3( )1 pt for answer

d. If the function w is given by

�

w x( ) = xf x( ) and ′ w 4( ) = 9, find ′ f 4( ). (2)

′w x( ) = x ′f x( ) + f x( )′w 4( ) = 4 ′f 4( ) + f 4( ) = 9⇒ 4 ′f 4( )− 5 = 9

4 ′f 4( ) = 14⇒ ′f 4( ) = 72

1 pt for ′w 4( )1 pt for answer


AB Calculus – Step-by-Step Solution Version 5. Trig Derivatives (Calculator allowed)

Let

�

f x( ) be given by

�

f x( ) = x + sin x which is defined on [0, 2π].

a. Find all exact values of x for which

�

′ f x( ) =1.5. (3)

′f x( ) = 1+ cos x = 1.5

cos x = 0.5 ⇒ cos x = 0.5 ⇒ x = π3

, x = 5π3

1 pt for ′f x( )2 pts for answers

b. If

�

g x( ) =f x( )x

, find the equation to the tangent line to g at

�

x = π2

. (3)

g x( ) = x + sin xx

= 1+ sin xx

g π2

⎛⎝⎜

⎞⎠⎟ = 1+

1π2

= 1+ 2π

′g x( ) = xcos x − sin xx2

⇒ ′gπ2

⎛⎝⎜

⎞⎠⎟ =

π2 0( )−1

π2( )2

= − 4π2

y − 1+ 2π

⎛⎝⎜

⎞⎠⎟ = − 4

π2x − π

2⎛⎝⎜

⎞⎠⎟

1 pt for g π2

⎛⎝⎜

⎞⎠⎟

1 pt for ′g π2

⎛⎝⎜

⎞⎠⎟

1 pt for equation

c. If

�

h x( ) = csc x , find all values of x on [0, 2π] where

�

′ f x( ) = ′ h x( ). (3)

h x( ) = 1sin x

⇒ ′h x( ) = −cos xsin2 x

1+ cos x = −cos xsin2 x

sin2 x 1+ cos x( ) = −cos x⇒ sin2 x 1+ cos x( ) + cos x = 0x = 2.032, x = 4.251

1 pt for setting student’s ′f x( ) = ′h x( )2 pt for answers


AB Calculus – Step-by-Step Solution Version 6. Linear Approximation

A line is tangent to the graph of

�

f x( ) = 25 − x2 at point P, as shown in the figure above.

a. Show that the x-coordinate of point P is 3. Explain your reasoning. (3)

Point P 3,16( )

′f x( ) = mparabola = −2x mline =34 −160 − 3

= −6

′f x( ) = −2 3( ) = −6At x = 3, the slope of the tangent line to the parabola = slope of the line

1 pt for ′f 3( )1 pt for slope of line1 pt for explanation

b. Find the equation of the line. (1)

m = −2 3( ) = −6y − 34 = −6x or y = 34 − 6x 1 pt for equation

c. Show that the difference between

�

f 3+ a( ) and the linear approximation to

�

f x( ) at

�

x = 3+ a where a is a constant gives the same value as the difference between

�

f 3− a( ) and the linear approximation to

�

f x( ) at

�

x = 3− a . (5)

At x = 3+ aApproximation = 34 − 6 3+ a( ) = 16 − 6a

f 3+ a( ) = 25 − 3+ a( )2 = 16 − 6a − a2

Difference = 16 − 6a − 16 − 6a − a2( ) = a2

At x = 3− aApproximation = 34 − 6 3− a( ) = 16 + 6a

f 3− a( ) = 25 − 3− a( )2 = 16 + 6a − a2

Difference = 16 + 6a − 16 + 6a − a2( ) = a2

1 pt for approximation at 3+ a1 pt for f 3+ a( )1 pt for approximation at 3− a1 pt for f 3− a( )1 pt for showing differences are the same


AB Calculus – Step-by-Step Solution Version 7. Chain Rule/Trig

x

f x( )

! f x( )

g x( )

! g x( )

1

3!4

2 0

1a

2 8 -6 3 -4

3 1 5 4

12

4 5

2a2 9 3

The functions f and g are differentiable for all real numbers x. The table above gives values of the function and their first derivatives at selected values of x with a being a constant.

a. If

h x( ) = sin f x( )( ) , write an equation of the line tangent to h at the point where x = 1. (3)

h 1( ) = sin f 1( ) = sin 3!4

"#$

%&' =

22

(h x( ) = cos f x( )( ) ) (f x( )* (h 1( ) = cos f 1( )( ) ) (f 1( ) = cos 3!4

"#$

%&' 2 = + 2

22( ) = +1

y + 22

= +1 x +1( ) or y = +x +1+ 22

1 pt for h 1( )1 pt for !h 1( )1 pt for equation

b. If

r x( ) = 1g 2x( )

, find

! r x( ) at x = 2. (2)

r x( ) = g 2x( )!" #$%1 2

& 'r x( ) = %12

g 2x( )!" #$%3 2

( 'g 2x( ) (2 = % 'g 2x( )g 2x( )!" #$

3 2

'r 2( ) = % 'g 4( )g 4( )!" #$

3 2 =%393 2

= %327

= %19

1 pt for !r x( )1 pt for !r 2( )

c. Find the value(s) of a if the tangent lines to

f g x( )( ) and

g f x( )( ) are perpendicular at x = 3. (4)

!f g x( )( ) " !g x( )#$ %& " !g f x( )( ) " !f x( )#$ %& = '1 at x = 3

!f g 3( )( ) " !g 3( )#$ %& " !g f 3( )( ) " !f 3( )#$ %& = '1

!f 4( ) " 12

#$(

%&)" !g 1( ) "5#$ %& = '1

2a2 12

*+,

-./

1a

*+,

-./ 5( ) = '1

5a = '10 a = '15

1 pt for derivative of f g x( )( ) at x = 3

1 pt for derivative of g f x( )( ) at x = 3

1 pt for m1( ) m2( ) = !1 or m2 =!1m1

1 pt for answer


AB Calculus – Step-by-Step Solution Version 8. Chain Rule Parts a, b, and c all refer to

�

f x( ) , given by

�

f x( ) = x2 − x − 6 which is defined on [0, 6].

a. Write an equation of the line tangent to f at the point where x = 4. (3)

f 4( ) = 42 − 4 − 6 = 6′f x( ) = 2x −1⇒ ′f 4( ) = 2 4( )−1= 7y − 6 = 7 x − 4( ) or y = 7x − 22

1 pt for f 4( )1 pt for ′f x( )1 pts for equation

b. If

�

g x( ) = f x( )[ ]2 , write an equation of any horizontal tangent lines to g. Show how you arrive at your answer. (3)

′g x( ) = 2 f x( ) ′f x( ) = 2 x2 − x − 6( ) 2x −1( ) = 2 x − 3( ) x + 2( ) 2x −1( ) = 0

x = 3,−2, 12

On domain, x = 3, 12

g 3( ) = 0 ⇒ y = 0 g 12

⎛⎝⎜

⎞⎠⎟ = 1

4− 1

2− 6⎛

⎝⎜⎞⎠⎟

2

= − 254

⎛⎝⎜

⎞⎠⎟

2

⇒ y = 62516

1 pt for ′g x( )2 pts for equation−1 for extraneous equations

c. If

�

h x( ) = 1f 2x( )

, find all values of x where the tangent lines to h are either horizontal or do not exist on

the interval [0, 6]. Show how you arrive at your answer. (3)

h x( ) = f 2x( )⎡⎣ ⎤⎦−1= 1

4x2 − 2x − 6

′h x( ) = − 8x − 2( )4x2 − 2x − 6( )2 =

−2 4x −1( )2 2x2 − x − 3( )⎡⎣ ⎤⎦

2 =−2 4x −1( )

2x − 3( ) x +1( )⎡⎣ ⎤⎦2

On 0,6[ ], horizontal tangent line at x = 14

and tangent line doesn’t exist at x = 32

.

1 pt for ′h x( )1 pt for horizontal1 pt for no tangent line−1 for extraneous values


AB Calculus – Step-by-Step Solution Version 9. Derivatives with ln/exponentials

Let

�

f x( ) = e + ln xx2

.

a. Find the average rate of change of f from x = 1 to x = e. (2)

f e( )− f 1( )e−1

=

e+ lnee2 − e+ ln1

12

e−1

=

e+1e2 − e

e−1= e+1− e3

e2 e−1( ) or e+1− e3

e3 − e2

1 pt for average rate of change formula1 pt for answer

b. Write an equation of the line tangent to f at x = 1. (3)

′f x( ) =x2 1

x⎛⎝⎜

⎞⎠⎟ − 2x e+ ln x( )

x4 =x 1− 2 e+ ln x( )⎡⎣ ⎤⎦

x4

′f x( ) = 1− 2 e+ ln x( )x3 ⇒ ′f 1( ) = 1− 2 e+ ln1( )

13 = 1− 2e

f 1( ) = e+ ln112 = e

Tangent line: y − e = 1− 2e( ) x −1( )

1 pt for ′f x( )1 pt for f 1( )1 pt for answer

c. Find the x-coordinate of the point on f at which the tangent line to f is horizontal. (2)

′f x( ) = 1− 2 e+ ln x( )x3

= 0

2 ln x = 1− 2e⇒ ln x = 1− 2e2

x = e1−2e2

1 pt for ′f x( ) = 01 pt for answer

d. Find

�

limx→0+

f x( ) and limx→∞

f x( ). (2)

limx→0+

f x( ) = −∞ or does not exist

limx→∞

f x( ) = 0

1 pt for limx→0+

f x( )1 pt for lim

x→∞f x( ) = 0


AB Calculus – Step-by-Step Solution Version 10. Chain Rule/Exponentials

Let

�

f x( ) be given by the function

�

f x( ) = ln x + 1x

⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

a. Show that

�

′ f x( ) = x2 −1x3 + x

. (2)

′f x( ) = 1

x + 1x

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

1− 1x2

⎛⎝⎜

⎞⎠⎟ =

xx2 +1

⎛⎝⎜

⎞⎠⎟

x2 −1x2

⎛⎝⎜

⎞⎠⎟

= x2 −1x x2 +1( ) =

x2 −1x3 + x

1 pt for ′f x( ) = 1

x + 1x

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

1− 1x2

⎛⎝⎜

⎞⎠⎟

1 pt for algebra

b. Find the x-coordinate of the point(s) on f at which the tangent line to f is horizontal. (1)

x2 −1= 0 ⇒ x = ±1 but x ≠ −1 so x = 1 1 pt for answer

c. Find the equation of the tangent line to

�

f x( ) at x = 2. (2)

′f 2( ) = 4 −18 + 2

= 310

y − ln 2.5( ) = 310

x − 2( )

1 pt for ′f 2( )1 pt for answer

d. If

�

g x( ) = e2 f x( ) , find

�

′ g 2( ). (3)

g x( ) = e2 f x( ) = e f x( )⎡⎣ ⎤⎦2= e

ln x+1x

⎛⎝⎜

⎞⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

= x + 1x

⎛⎝⎜

⎞⎠⎟

2

′g x( ) = 2 x + 1x

⎛⎝⎜

⎞⎠⎟ 1− 1

x2⎛⎝⎜

⎞⎠⎟ or ′g x( ) = 2 x2 +1( ) x2 −1( )

x3

′g 2( ) = 2 5( ) 3( )8

= 308

= 154

1 pt for g x( ) = x + 1x

⎛⎝⎜

⎞⎠⎟

2

1 pt for ′g x( )1 pt for answer

e. Show that

�

f x( ) and g x( ) have horizontal tangent lines at the same x-value(s). (1)

′g 1( ) = 2 2( ) 0( )1

= 0 1 pt for answer


AB Calculus – Step-by-Step Solution Version 11. Implicit Differentiation Consider the closed curve in the xy-plane given by

�

x2 − 6x + y3 −12y =11.

a. Show that

�

dydx

= 6 − 2x3y2 −12

. (2)

�

2x − 6 + 3y2 dydx

−12 dydx

= 0

dydx

3y2 −12( ) = 6 − 2x⇒ dydx

=6 − 2x3y2 −12

�

1 pt for implicit differentiation1 pt for verification

b. Write an equation for the line tangent to the curve at the point (6, -1). (2)

�

dydx 6,−1( )

= 6 −123−12

= −6−9

= 23

Tangent line : y + 1 = 23x − 6( ) or y = 2

3x − 5

�

1 pt for dydx

= 23

1 pt for equation

c. Find the coordinates of all points on the curve where the line tangent to the curve is vertical. (3)

Tangent lines occur when 3y2 −12 = 0 or y = ±2y = 2 : x2 − 6x + 8 − 24 = 11⇒ x2 − 6x − 27 = 0 x + 3( ) x − 9( ) = 0 ⇒ x = −3, 9 so vertical tangents at −3,2( ), 9,2( )y = −2 : x2 − 6x − 8 + 24 = 11⇒ x2 − 6x + 5 = 0 x −1( ) x − 5( ) = 0 ⇒ x = 1, 5 so vertical tangents at 1,− 2( ), 5,−2( )

�

1 pt for 3y2 = 121 pt for −3,2( ), 9,2( )1 pt for 1,−2( ), 5,−2( )2 pts maximum for only 2 of the 4 points

d. Show that it is impossible for this curve to have a horizontal tangent along the line y = 4. (2)

�

Horizontal tangents occur when 6 − 2x = 0 or x = 3At 3,4( ) : 9 −18 + 64 − 48 = 73− 66 = 7

So 3,4( ) is not on x 2 − 6x + y3 −12y = 11

�

1 pt for 6 − 2x = 01 pt for showing 3,4( ) not on curve


AB Calculus – Step-by-Step Solution Version 12. Implicit Differentiation/Linear Approximation (Calculator allowed)

Consider the closed curve in the xy-plane given by

�

2x2 − xy + y3 + x = 9.

a. Show that

�

dydx

= y − 4x −13y2 − x

. (2)

4x − x dydx

− y + 3y2 dydx

+1= 0

dydx

3y2 − x( ) = y − 4x −1⇒ dydx

= y − 4x −13y2 − x

1 pt for implicit differentiation1 pt for answer

b. Find equation(s) of all tangent lines to the curve at y = 1. (4)

c. There is a number k so that the point (2.1, k) is on the curve. Using the tangent line found in part b, approximate the value of k. (1)

y = 17 − 8 2.1( ) = 0.2k ≈ 0.2

�

1 pt for k ≈−1 if not ≈

d. Write an equation that can be solved to find the actual value of k so that the point (2.1, k) is on the curve. (1)

2x2 − xy + y3 + x = 9

2 2.12( )− 2.1k + k 3 + 2.1= 9

or k 3 − 2.1k +1.92 = 0

1 pt for equation

e. Solve the equation in part d) for k. (1)

k = −1.782

�

1 pt for k

2x2 − xy + y3 + x = 9 2x2 − x +1+ x = 92x2 = 8 ⇒ x = ±2

x = 2 : dydx

= 1− 8 −13− 2

= −8

y −1= −8 x − 2( ) ory = 17 − 8x

x = −2 : dydx

= 1+ 8 −13+ 2

= 85

y −1= 85x + 2( )

1 pt for dydx

at x = 2

1 pt for line at x = 2

1 pt for dydx

at x = −2

1 pt for line at x = −2


AB Calculus – Step-by-Step Solution Version 13. Implicit Differentiation/Exponentials Curves f and g are given by the equations below as shown in the figure to the right.

Curve f:

�

x2 − 9ln 2y −1( ) + y2 = 5 Curve g:

�

x2 + ey2−1 − y = 4

a. For curve f, show that

�

dydx

= 2xy − x9 − 2y2 + y

(2)

�

2x − 92y −1

2 dydx

⎛ ⎝ ⎜

⎞ ⎠ ⎟ + 2y

dydx

= 0

dydx

182y −1

− 2y⎛

⎝ ⎜

⎞

⎠ ⎟ = 2x⇒ dy

dx= 2x

182y −1

− 2y

dydx

=2x 2y −1( )

18 − 2y 2y −1( )= 4xy − 2x18 − 4y2 + 2y

= 2xy − x9 − 2y2 + y

�

1 pt for implicit1 pt for algebra

b. Show that horizontal tangents to curve f must occur along the y-axis. (2)

�

dydx

= 0 when 2x 2y −1( ) = 0

For curve f , y must be a number greater than 12

So 2x = 0 ⇒ x = 0 so horizontal tangents occur along y − axis

�

1 pt for 2x 2y −1( ) = 01 pt for y > 0

c. For curve g, find

�

dydx

. (2)

�

2x + ey2−1 2y( ) dy

dx− dydx

= 0

dydx1− 2yey

2−1( ) = 2x⇒ dydx

= 2x1− 2yey

2−1

�

1 pt for implicit differentiation1 pt for algebra

d. Show that the line tangent to f is the same as the line normal to curve g at (2, 1). (3)

�

Curve f : dydx 2,1( )

=2 2( ) 1( ) − 2

9 − 2 12( ) + 1= 4 − 2

9 − 2 + 1= 2

8= 1

4

Curve g : dydx 2,1( )

=2 2( )

1− 2 1( )e1−1 = 41− 2

= −4 so normal line has slope 14

Since the slopes are the same and they pass through the same point, the lines are the same.

�

1 pt for dydx 2,1( )

for f

1 pt for dydx 2,1( )

for g

1 pt for explanation


AB Calculus – Step-by-Step Solution Version 14. Continuity/Differentiability

Let

�


�

f x( ) =9 − 4mx − 1− x( )2 if x ≤1

m2x − n if x > 1

⎧ ⎨ ⎪

⎩ ⎪ where m and n are constants and

�

m ≠ 0.

a. Write an expression for n if f is continuous at x = 1. (1)

�

limx→1−

f x( ) = limx→1+

f x( )

9 − 4m = m2 − n⇒ n = m2 + 4m − 9

�

1 pt for expression

b. Show that f cannot be continuous at x = 1 if

�

n ≤ −14 . (2)

�

−14 = m2 + 4m − 9 ⇒ m2 + 4m + 5 = 0

m = −4 ± 16 − 202

= −4 ± 2i2

For any values of n ≤ −14, m will be imgainary

�

1 pt for quadratic equation1 pt for explanation involving imaginary−1 if explanation only involves n = −14

c. If f is differentiable at x = 1, find the values of m and n. Show your reasoning. (3)

�

′ f x( ) =−4m − 2 1− x( ) −1( ) if x ≤1

m2 if x > 1

⎧ ⎨ ⎪

⎩ ⎪

limx→1−

′ f x( ) = limx→1+

′ f x( )

−4m = m2 ⇒ m2 + 4m = 0 ⇒ m = −4n = 16 −16 − 9 ⇒ n = −9

�

1 pt for ′ f x( )1 pt for n1 pt for m

d. Using the values of m and n found in part c), determine values of x (if any) that will make

�

′ f x( ) differentiable. Show your reasoning. (3)

�

′ f x( ) =16 + 2 1− x( ) if x ≤ 116 if x > 1⎧ ⎨ ⎩

′ ′ f x( ) =−2 if x ≤ 10 if x > 1⎧ ⎨ ⎩

limx→1−

′ ′ f x( ) ≠ limx→1+

′ ′ f x( ) as − 2 ≠ 0

So ′ f x( ) is differentiable at all x - values except x = 1

�

1 pt for ′ f x( )1 pt for ′ ′ f x( )1 pt for answer


AB Calculus – Step-by-Step Solution Version 15. Continuity/Differentiability w/Trig

Let

�


�

f x( ) =g x( ) + a if x ≤ 03− bcos x if x > 0

⎧ ⎨ ⎩

where a and b are constants and

�

g x( ) = 1− x2 .

a. Determine if

�

g x( ) is differentiable at x = 1. Justify your answer. (3)

g x( ) = 1− x2 = x2 −1

g x( ) =x2 −1 if x ≥1

− x2 −1( ) if x <1

⎧⎨⎪

⎩⎪limx→1−

g x( ) = 0 limx→1+

g x( ) = 0

so g x( ) is continuous at x =1

′g x( ) =2x if x ≥1−2x if x <1

⎧⎨⎪

⎩⎪limx→1−

′g x( ) = 2 limx→1+

′g x( ) = −2

so g x( ) is not differentiable at x =1

�

1 pt for writing g x( ) as a piecewise function1 pt for cont. with limits1 pt for diff. with limits

b. Show that

�

f x( ) is differentiable at x = 0 if a =1 and b = 1. (3)

f x( ) =x2 if x < −11− x2 +1 if −1≤ x ≤ 03− cos x if x > 0

⎧

⎨⎪

⎩⎪

limx→0−

f x( ) = 2 limx→0+

f x( ) = 3−1= 2

So f is continuous at x = 0

′f x( ) = −2x if x ≤ 0sin x if x > 0⎧⎨⎩

limx→0−

′f x( ) = 0 limx→0+

′f x( ) = 0

so f is differentiable at x = 0

�

1 pt for using 1− x2

1 pt for cont. with limits1 pt for diff. with limits

c. Find a relationship between a and b in order for

�

f x( ) to be continuous at x = 0. (1)

f x( ) =x2 if x < −11− x2 +1 if −1≤ x ≤ 03− cos x if x > 0

⎧

⎨⎪

⎩⎪

limx→0−

f x( ) =1+ a limx→0+

f x( ) = 3− b

so 1+ a = 3− b or a + b = 2

�

1 pt for a + b = 2

d. Find a relationship between a and b in order for

�

f x( ) to be differentiable at x = 0. (2)

′f x( ) =2x if x < −1−2x if −1≤ x ≤ 0bsin x if x > 0

⎧⎨⎪

⎩⎪

limx→0−

′f x( ) = 0 limx→0+

′f x( ) = 0

so f is differentiable for any value of b if a + b = 2

�

1 pt for ′ f x( )1 pt for answer.

ab calculus – step-by-step solution version 1. limits · © 2012 - 2 - illegal to post this...

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