SOLUTIONS

Math 220 GW 7 SOLUTIONS

1. Using the limit definition of derivative, find the derivative function, f (x), of the following functions. Show all your beautiful algebra.

(a) f (x) = 2x

f (x + h) - f (x)

2(x + h) - 2x

lim

= lim

h0

h

h0

h

2x + 2h - 2x

= lim

h0

h

2h = lim

h0 h

2.

(b) f (x) = -x2 + 2x

f (x + h) - f (x)

-(x + h)2 + 2(x + h) + x2 - 2x

lim

= lim

h0

h

h0

h

-x2 - 2xh - h2 + 2x + 2h + x2 - 2x

= lim

h0

h

-2xh - h2 + 2h

= lim

h0

h

= lim (-2x - h + 2)

h0

= -2x + 2.

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2. You are told f (x) = 2x3 - 4x, and f (x) = 6x2 - 4. Find f (3) and f (-1) and explain, in words, how to interpret these numbers.

f (3) = 6(3)2 - 4 = 50. f (1) = 6(1)2 - 4 = 2.

These are the slopes of f (x) at x = 3 and x = 1. Both are positive, thus f is increasing at those points. Also, 50 > 2, so f is increasing faster at x = 3 than at x = 1. Example Find the derivative of f (x) = 3/x2.

f (x + h) - f (x)

f (x) = lim

h0

h

=

lim

3 (x+h)2

-

3 x2

h0

h

=

lim

x2 x2

3 (x+h)2

-

3 x2

(x+h)2 (x+h)2

h0

h

3x2-3(x+h)2

= lim

h0

x2(x+h)2 h

1

3x2 - 3(x + h)2 1

= lim

h0

x2(x + h)2

h

3x2 - 3(x2 + 2xh + h2)

= lim

h0

hx2(x + h)2

3x2 - 3x2 - 6xh + h2

= lim

h0

hx2(x + h)2

h(-6x + h)

=

lim

h0

hx2(x

+

h)2

h -6x + h

=

lim

h0

h

x2(x

+

h)2

-6x + h

=

lim

h0

x2(x

+

h)2

-6x + 0 = x2(x + 0)2

-6x = x4

-6 = x3

3. For the following questions consider f (x) = 4/x.

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(a) Find the derivative f (x).

lim

f (x + h) - f (x)

=

lim

4 x+h

-

4 x

h0

h

h0 h

4x-4(x+h)

= lim x(x+h)

h0

h

4h = lim

h0 hx(x + h)

4 = lim

h0 x(x + h)

4 = x2 .

(b) Find and interpret f (5).

44

f

(5)

=

52

=

. 25

The

slope

of

f

at

x

=

5

is

4 25

.

(c) When x = 5, is the graph of f (x) increasing, decreasing, or nei-

ther? Explain why.

Since

4 25

is

positive,

f (x)

is

increasing

at

x

=

5.

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 4. Let f (x) = x - 5

(a) Find the equation of the secant line that goes through the graph

when x = 9 and x = 14.

First, we find the slope of the secant line:

f (14) - f (9) 14 - 5 - 9 - 5 3 - 2 1

=

=

=.

14 - 9

5

55

Then, we use point-slope formula for a line using the point (9, f (9)) =

(9, 2):

1 y - 2 = (x - 9).

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(b) Find the equation of the tangent line to the graph of f (x) when

f (9 + h) - f (9)

x = 9.(Hint: Calculate lim

to find the slope of

h0

h

the tangent line at x = 9)

Finding the tangent line, we need the tangent slope:

f (9 + h) - f (9)

(9 + h) - 5 - 2

lim

= lim

h0

h

h0

h

4+h-2 4+h+2

= lim

h0

h

4+h+2

4+h-4 = lim

h0 h( 4 + h + 2)

h = lim

h0 h( 4 + h + 2)

1 = lim

h0 4 + h + 2

1 =.

4

Using point-slope formula using the point (9, f (9)) = (9, 2):

1 y - 2 = (x - 9).

4

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5. Match the 3 functions below to their derivatives. Do this by considering when a function is increasing its derivative is positive, when a function is decreasing its derivative is negative, and when a function has a min/max its derivative is 0. Functions

Possible Derivatives

Function 1--Derivate 2; Function 2--Derivate 1; Function 3--Derivate 3.

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6. Consider the graph below to be the graph of f (x), the DERIVATIVE of some unknown function f (x). Use this graph to answer the following questions.

(a) What is the slope of the tangent line to the graph of f (x) when x = 1? f (1) = 1.

(b) Knowing that the function f (x) can only have a min/max at the point x = a if f (a) = 0, what are the possible mins/maxes of f (x)? x = 1.6, 3.2, 4.7.

(c) Knowing the the function f (x) can only be increasing at the point x = a if f (a) is positive, on what intervals is f (x) increasing? [-1, 1.6] [3.2, 4.7]. Note: we don't know if it extends below -1.

(d) Knowing that the function f (x) can only be decreasing at the point x = a if f (a) is negative, on what intervals is f (x) decreasing? [1.6, 3.2] [4.7, 5]. Note: we don't know if it extends past 5

(e) Find all points x = a where the tangent line to f (x) is horizontal. These are the same as part (b): x = 1.6, 3.2, 4.7.

(f) Sketch a graph of f (x).

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