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(10) 1. Suppose f(x)=2x

2

3x. Use the denition of derivative to nd f

0

(x).

f

0

(x)=lim

h!0

f(x+h)f(x)

h

= lim

h!0

2(x+h)

2

3(x+h)(2x

2

3x)

h

= lim

h!0

2x

2

+4xh +2h

2

3x3h2x

2

+3x

h

= lim

h!0

4xh +2h

2

3h

h

= lim

h!0

4x+2h3=4x3

(9) 2. Find an equation for the line tangent to the graph of y=

p

x+2x

2

at the point where

x=1.

y=x

1=2

+2x

2

,soy

0

=

1

2

x

1=2

+4x. At x= 1, the value yis 3 and the value of y

0

is 9=2.

Thus an equation for the tangent is

y3=

9

2

(x1):

(12) 3. Assume that the functions u(x) and v(x) are dened and dierentiable for all real

numbers x. The following data is known about u,v, and their derivatives.

x u(x)v(x)u

0

(x)v

0

(x)

2 3 4 1 2

3 2 1 3 1

4 1 3 0 2

Dene f(x)=u(x)

2

+2v(x) and g(x) = v(x)=u(x). Answer the following, giving a brief

explanation of how the answers were obtained.

a) What is f

0

(2)?

Since the chain rule had not been covered when the test was given, to dierentiate u(x)

2

we have to write it as u(x)u(x) and use the product rule.

f

0

(x)=u(x)u

0

(x)+u

0

(x)u(x)+2v

0

(x)=2u(x)u

0

(x)+2v

0

(x):

Thus

f

0

(2) = 2(3)(1) + 2(2) = 2:

b) What is g

0

(3)?

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g

0

(x)=

u(x)v

0

(x)v(x)u

0

(x)

u

2

(x)

Thus

g

0

(3) =

2(1) 1(3)

2

2

=

5

4

:

c) What can be said about the number and location of solutions to the equation f(x)=6:5

with xin [2;4]?

From the table, we have f(2) = 17, f(3) = 6, and f(4) = 7. By the Intermediate Value

Theorem, there is at least one solution to the equation f(x)=6:5 in the interval [2;3] and

at least one in the interval [3;4]. Thus the total number of solutions is at least 2.

(12) 4. Suppose that the function f(x)isdescribed by

f(x)=

x+Bif x<1

Ax +3 if x1

:

a) Find Aand Bso that f(x) is continuous for all numbers and f(1) = 0. Briey explain

your answer.

The only place that fmight not be continuous is at x=1,where the denition changes.

Now f(1) = A+ 3 while lim

x!1

f(x)=1+B. If fis to be continuous at x=1,wemust

have A+3=1+B.

The value of f(1) is 1+B, whichmust be 0. This gives B=1.Substituting this value

in the previous equation, we get A=1.

b) Sketch y=f(x) on the axes given for the values of Aand Bfound in a) when xis in

the interval [2;2].

–4

–3

–2

–1

1

2

3

4

–4–3–2–1 1234

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