SAMPLE MIDTERM 2 - SOLUTIONS:

1. (24 pts) Evaluate the derivatives of the following functions.

NO NEED TO SIMLIFY!

(a) (8 pts) y = tan x5 + 4)

y ' =1

2 tan x 5 + 4 )

sec2 x 5 + 4 5 x 4

(b) (8 pts) y = xcos x

Since

y = x cos x =eln x cos x = ecos x ln x ,

y ' =ecos x ln x ddx

(cos x ln x) =

=ecos x ln x - sin x ln x +cos xx

=

= x cos x - sin x ln x +cos xx

(II) (8 pts) Use implicit differentiation to finddy

dx.

x2 - y2 = 4

ddx

x 2 - y 2 =ddx

(4)

2 x - 2 ydy

dx= 0

x - ydy

dx= 0

ydy

dx= x

dy

dx=xy

2. (20 pts)

A boat sails directly toward a 200 - meter skyscraper that stands on the edgeof a harbour. The angular size θ of the building is the angle formed by linesfrom the top and bottom of the building to the observer (see figure).

(a) Express the angle θ as the function of x, the distance of the boatfrom the building.

Since tan θ =200

x, it follows that

θ (x) = tan-1 200

x

The boat is sailing directly toward the skyscraper at 3

when the boat is x = 500 m from the building.

2 SOLUTIONS - SAMPLE MIDTERM 2.nb

(b) The boat is sailing directly toward the skyscraper at 3 m s.

Finddθ

dtwhen the boat is x = 500 m from the building.

Since tan θ =200

x, it follows that

ddt

(tan θ ) =ddt

200

xand, therefore

sec2 θdθ

dt= -

200

x 2dx

dt

We have to evaluate sec2 θ at the moment when x = 500 m.

sec θ =5002 + 2002

500= 100

5 2 + 2 2

500=

295

By plugging in x = 500 ,dxdt

= -3, and sec θ =295

, we obtain

295

2 dθdt x=500

= -200

5002(-3) ,

dθdt x=500

= -200

5002(-3)

2529

and

dθdt x=500

=2100

329

=6

2900rad /sec

OR

SOLUTIONS - SAMPLE MIDTERM 2.nb 3

Since θ = tan-1 200x

, it follows thatddt

θ =ddttan-1 200

x, and, therefore

dθdt

=ddx

tan-1 200x

dxdt

dθdt

=1

1 + 200x 2-200

x 2dxdt

dθdt

=1

1 + 2002

x 2

-200

x 2dxdt

=-200

x 2 + 2002dxdt, and, by plugging in x = 500, and

dxdt

= -3,

we obtain thatdθdt x=500

=-200

5002 + 2002(-3) =

62500 + 400

=6

2900rad /sec

3. (20 pts)

Sketch (neatly !) the graph of a function f satisfying all of the following conditions :

(a) f (0) = 10 , f (-6) = 10

(b) limx→-2

f (x) = + ∞ , limx→+∞

f (x) = 0,

(c) f ' (x) > 0 on (-∞ , -12) and (-6, -2),

(d) f ' (x) < 0 on (-12, -6), and (-2, +∞ ),

(e) f '' (x) > 0 on (-8, -2) and (-2, +∞ ),

(f) f '' (x) < 0 on (-∞ , -8) .

4 SOLUTIONS - SAMPLE MIDTERM 2.nb

y = f ( x )

-20 -18 -16 -14 -12 -10 -8 -6 -4 -2 2 4 6 8 10 12 14 16 18 20x

-20

-15

-10

-5

5

10

15

20y

4. (18 pts) MULTIPLE CHOICE!!!

The derivative of function f is given and a table of values for

g (x) and g' (x) is shown below.

Suppose that g is a one - to - one function and g-1 (x) is its inverse.

f' (x) =x

x2 + 1,

x g (x) g' (x)1 2 42 3 53 4 1

SOLUTIONS - SAMPLE MIDTERM 2.nb 5

CIRCLE ALL THE CORRECT ANSWERS IN EACH PART.

(I) Find all critical points of f .

(a) x = -2 ; (b) x = -1 ; (c) x = 2 ; (d) x = 0 ;

(e) x = 1 ; (f )NONE OF THE PREVIOUS ANSWERS.

(II) On what interval (s) is f increasing?

(a) (-2, +∞); (b) (-1, +∞) ; (c) (-1, 1) ; (d) (-1, 0) ;

(e) (-∞, 0 ) ; (f ) (0, +∞ ) (g)NONE OF THE PREVIOUS ANSWERS.

4. (CONTINUED)

(III) The function f has a local maximum at

(a) x = -2 ; (b) x = -1 ; (c) x = 0 ; (d) x = 1 ;

6 SOLUTIONS - SAMPLE MIDTERM 2.nb

(e) x = 2 ; (f ) The function f has no local maxima.

(IV)d

dxf (g (x)) at x = 1.

(a) 2 ; (b)25; (c)

85

(d) 1 ;

(e) 0 ; (f )NONE OF THE PREVIOUS ANSWERS.

f ' (g (1)) g ' (1) = f ' (2) (4) =2

2 2 + 14 =

85

(V) g-1 (2)

(a) 3 ; (b) 1 ; (c)13; (d) 4;

(e) DOES NOT EXIST ; (f )NONE OF THE PREVIOUS ANSWERS.

Check the table : g (1) = 2 , therefore g -1 (2) = 1 .

(VI) d

dxg-1 (x) at x = 2.

(a) 1 ; (b) 4 ; (c)14

; (d)15

;

(e) 0 ; (f )NONE OF THE PREVIOUS ANSWERS.

ddx

g -1 (2) =1

g ' (b)where b = g -1 (2). Therefore,

ddx

g -1 (2) =1

g ' (1)=14

, AND NO PARTIAL CREDIT WILL BE

graph of is shown in the

SOLUTIONS - SAMPLE MIDTERM 2.nb 7

5. (18 pts) EXPLANATION IS NOT REQUIRED, AND NO PARTIAL CREDIT WILL BE GIVEN.

The graph of f' derivative of f is shown in the figure.

Assume that the function f is continuous on (-∞, +∞).

y = f ' ( x )

-2 -1 1 2 3 4x

-1

1

2

3

4

y = f'(x)

Use the given graph of f' to answer the following questions about f :

(a) (4 pts) On what interval (or intervals) is f increasing?

(-∞, -1) and (1, 4) ( because f ' is positive there).

(b) (4 pts) Find the critical points of f.

x = -1, x = 1 , (because f ' is equal to zero there)

and x = 4 (because f ' does not exist there)

(c) (4 pts) Which critical point (or points) correspond to local maxima?

8 SOLUTIONS - SAMPLE MIDTERM 2.nb

x = -1, and = 4 (because f ' changes sign from positive to negative

as x increases through these points )

(d) (3 pts) On what interval (or intervals) is f concave down?

(-∞, 0) (because f ' is decreasing there)

(e) (3 pts) At what point (or points) does f have an inflection point?

x = 0 , (because f changes concavity there)

SOLUTIONS - SAMPLE MIDTERM 2.nb 9