Exam 1 Solutions

[Pages:5]Exam 1 Solutions

1. (25 points) Find each limit. a. limx3(x2 - 3x + 7)

limx3(x2 - 3x + 7) = (3)2 - 3(3) + 7 = 7

b.

limx2

x-2 x+2

limx2

x-2 x+2

=

(2)-2 (2)+2

=

0 4

=

0

c.

limx1

x2 -1 x2 +x-2

Simply plugging gives '0/0' which is indeterminate. So, we simplify first:

limx1

x2 -1 x2 +x-2

=

limx1

(x-1)(x+1) (x-1)(x+2)

=

limx1

x+1 x+2

=

(1)+1 (1)+2

=

2 3

d.

limx

x-2x3 x3 +2x2 +1

For 'limits at infinity' of 'rational functions' (fractions):

divide everything by the highest power of x in the denominator:

limx

x-2x3 x3 +2x2 +1

=

limx

x x3

-

2x3 x3

x3 x3

+

2x2 x3

+

1 x3

=

limx

1 x2

-2

1+

2 x

+

1 x3

=

0-2 1+0+0

=

-2

e.

limx-

x2 +1 x5 -7

Same trick as in d) gives 0

1

2. (10 points) Estimate the limit by making a table of values. Use three values to the left, three to the right.

limx7

x2 +6x-91 x-7

For example, plug in 6.9, 6.99, 6.999, and 7.1, 7.01, 7.001.

3. (15 points) Use the graph to fill in the blanks below. (COMING SOON)

limxa- f (x) = limxb- f (x) = limxc- f (x) = limx f (x) =

limxa+ f (x) = limxb+ f (x) = limxc+ f (x) = limx- f (x) =

limxa f (x) = limxb f (x) = limxc f (x) = limxc f (a) =

f (a) = f (b) = f (c) =

2

4. (10 points) Use the limit definition to find the derivative of f (x) = x2 - 2x.

f (x + h) - f (x)

f (x) = lim

h0

h

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

= lim

h0

h

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

= lim

h0

h

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

= lim

h0

h

2xh + h2 - 2h

= lim

h0

h

h(2x + h - 2)

= lim

h0

h

= lim (2x + h - 2)

h0

= 2x + (0) - 2

= 2x - 2

5. (15 points) Use the 'shortcut rules' to find the derivative of each function. a. f (x) = x2 + 7x - 2

f (x) = 2x + 7

b. f (x) = x + 2

f (x)

=

x1 2

+2

=

f

(x) =

1 2

x-

1 2

+0

c.

f (x)

=

x 2

+

2 x

f (x)

=

1 2

x

+

2x-1

=

f

(x)

=

1 2

-

2x-2

3

6. (18 points) Consider the function f (x) = x3 - 9x + 6. a. Find the equation of the tangent line to f at x = 2 (by any method).

The tangent line to f at x = a is the line through (a, f (a)) with slope f (a): ? (x1, y1) = (2, f (2)) = 2, (2)3 - 9(2) + 6 = (2, -4) ? f (x) = 3x2 - 9 = m = f (2) = 3(2)2 - 9 = 3

Since m = 3, we know this tangent line looks like: y = 3x + b

Since the line goes through (x, y) = (2, -4), we have (-4) = 3(2) + b

which says b = -10. Thus,

y = 3x - 10

b. Find all points x where the tangent line to f is horizontal. The tangent line to f at x has slope f (x) and, thus, is horizontal if and only if

f (x) = 0

For f (x) = x3 - 9x + 6, we have ...

4

f (x) = 0 3x2 - 9 = 0 3(x2 - 3) = 0 x2 - 3 = 0

x2 = 3

x = ? 3

7. (7 points) Let f (x) = x7. (As always, justify your answers!) a. Find f (1).

f (x) = 7x6 = f (1) = 7.

b. Find the following limit.

(1 + h)7 - 1

lim

h0

h

For any function f , the definition of the derivative at x = a is

f (a + h) - f (a)

f (a) = lim

h0

h

Thus, the above limit is simply the definition of f (1). (Verify this!). Thus,

(1 + h)7 - 1

lim

= f (1) = 7

h0

h

5

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