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Name

MATH 105 FINAL

Key

Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.

PROBLEM 1 Please answer the following true or false. If false, explain why or provide a counter example. If true, explain why.

A) If f(x) is a differentiable function that passes through the origin such that f '(x) > 2 for all x, then f(5) cannot equal 10.

Solution

True, by the Mean Value Theorem, if f(5) = 10 then there is a c between 0 and 5 with

                             f(5) - f(0)            10 - 0
            f '(c) =                         =                   = 2
                                5 - 0                    5

B) If f(x) is a continuous function such that f '(0) = 2, f '(1) = 0, and f '(2) = -3 then f(x) has a relative maximum at x = 1.

Solution

False, The first derivative test needs that within a neighborhood of x = 1 all values to the left have positive derivative and all values to the right have negative derivative.

C) Suppose that h(x) = g '(x) and that f(x) and h(x) are continuous. Then if g(a) = g(b),

$ \int_a^b f(g(x)g'(x)dx = 0 $

Solution

True, use u-substitution with u = g(x), du = g '(x), then when x = a, u = g(a) and when x = b, u = g(a) also. Since the bottom and top limit are the same, the integral is zero.

PROBLEM 2 Find the derivative of

A. f(x) = x cos(x²)

Solution

We use the product and chain rule. Since the derivative of

cos(x²)

(2x)(-sin(x²)) = -2x sin(x²)

The product rule gives

f '(x) = (x)(-2x sin(x²)) + (1)(cos(x²)) = -2x² sin(x²) + cos(x²)

B.                      x² - 1
           f(x) =
                          x² + 1

Solution

Use the quotient rule

                             (x² + 1)(2x) - (x² - 1)(2x)
           f '(x) =
                                         (x² + 1)²

                    2x³ + 2x - 2x³ + 2x
           =
                            (x² + 1)²

                          4x
             =
                    (x² + 1)²

PROBLEM 3 Find the limit if it exist

$ \lim\limits_{x \to 1} \frac{1 - \sqrt{x}}{1 - x} $

Solution

Multiply numerator and denominator by the conjugate

$ \lim\limits_{x \to 1} \frac{1 - \sqrt{x}}{1 - x} \frac{1 + \sqrt{x}}{1 + \sqrt{x}} = \lim\limits_{x \to 1} \frac{1 - x} {(1 - x)(1 + \sqrt{x})} $

$ \lim\limits_{x \to 1} \frac{1}{1 + \sqrt{x}} = \frac{1}{2} $

PROBLEM 4 Evaluate the following integrals.

A) $ \int_1^3 (x - 1)\sqrt{x^2 - 2x + 6} dx $

Solution

Let u = x² - 2x + 6, du = (2x - 2)dx, (x - 1)dx = du/2

When x = 1, u = 5 and when x = 3, u = 9.

We get

$ \frac{1}{2} \int_5^9 u^{\frac{1}{2}} du = \frac{1}{2} [\frac{2}{3}u^{\frac{2}{3}}]_5^9$

$ = \frac{1}{3}[9^{\frac{3}{2}} - 5^{\frac{3}{2}}] = 9 - \frac{5^{\frac{3}{2}}}{3} $

B) $ \int (x^2 + 1)^2 + 3 dx $

Solution

First FOIL out and then integrate:

$ \int (x^4 + 2x^2 + 4) dx = \frac{1}{5}x^5 + \frac{2}{3}x^3 + 4x + C $

PROBLEM 5 (35 Points) You have a camera that rotates automatically positioned 400m from the space shuttle launch pad. When the space shuttle is 300m from the ground the shuttle is moving at 20 meters per second. How fast should your camera rotate at that instant?

Solution

First we sketch the picture and label variables

$Right triangle with base 400, height y = 300 and dy/dt = 20, and angle theta. There is a rocket ship on the top right vertex.$

Now use the trigonometry formula:

tan q = y/400

Now take the derivative of both sides with respect to t:

sec²q dq/dt = 1/400 dy/dt

Next plug in y = 300 and use the triangle to find sec²q. Note that this is a 300-400-500 triangle, so that

sec²q = (5/4)²

We have

(25/16)dq/dt = (1/400) 20 = 1/20

Hence

dq/dt =(1/20)(16/25) = 4/125 radians per second

PROBLEM 6 (35 Points) Use right sums with n = 200 to approximate the area under the curve y = 2x + 1, above the x-axis between x = 4 and x = 10.

Solution

We have

$ \sum_{i=1}^{200} [2(4 + \frac{6i}{20})+1]\frac{6}{200} $

$ = \frac{6}{200} \sum_{i=1}^{200} (9 + \frac{3i}{50}) = \frac{6}{200} \sum_{i=1}^{200} 9 + \frac{18}{10,000} \sum_{i=1}^{200} i $

$ = (\frac{6}{200})(9 \cdot 200) + (\frac{18}{10,000})(\frac{(200)(201)}{2}) $

$ = 54 + 36.18 = 90.18 $

PROBLEM 7 (35 Points) Let $ \int_{1 - x^2}^{10} ln(t^2 + 1) dt $ Find F '(x).

Solution

We use the chain rule, a property of integrals and the second fundamental theorem of calculus:

u = 1 - x² u' = -2x

$ F(u) = \int_{u}^{10} ln(t^2 + 1) dt = -\int_{10}^{u} ln(t^2 + 1) dt $

Putting this all together gives

F '(x) = (-2x)(-ln(u² + 1)) = 2x ln[(1 - x²)² + 1]

PROBLEM 8 (35 Points) Use the limit definition of the derivative to find the derivative of

f(x) = x² - 2x

Solution

$ f'(x) = \lim\limits_{h \to 0} \frac{[(x+h)^2 - 2(x+h)] - [x^2 - 2x]}{h} $

$ = \lim\limits_{h \to 0} \frac{[(x^2 + 2xh +h^2 -2x - 2h] - [x^2 - 2x]}{h} $

$ = \lim\limits_{h \to 0} \frac{2xh + h^2 - 2h}{h} = \lim\limits_{h \to 0} \frac{h(2x + h - 2)}{h} $

$ = \lim\limits_{h \to 0} (2x + h - 2) = 2x - 2$

PROBLEM 9 Let

                          1
        f(x) =
                       1 - x²

Determine any relative extrema, inflection points, intervals where y = f(x) is increasing, intervals where y = f(x) is concave up, and any asymptotes. Then use this information (not you calculator!) to graph the function.

Solution

We have

f '(x) = 2x(1 - x²) ^-2

Hence there is a critical point at (0,1). There are also vertical asymptotes at

x = 1 and x = -1

For x < -1, f '(x) < 0

For -1 < x < 0, f '(x) < 0

For 0 < x < 1, f '(x) > 0

For 1 < x, f '(x) > 0

Hence the graph is decreasing on (-, -1) U (-1, 0)

and increasing on (0, 1) U (1, )

By the first derivative test, there is a minimum at (0,1).

We have

f ''(x) = 2(1 - x²) ^-2 + (2x)(-2x)(-2)(1 - x²)^-3

2(1 - x²) ^-2 + (8x²)(1 - x²)^-3

= (1 - x²) ^-3[2 - 2x² + 8x²]

= (1 - x²) ^-3[2 + 6x²]

Since f ''(x) is never zero there are no inflection points. We can check for concavity:

When x < -1, f ''(x) < 0

When -1 < x < 1, f ''(x) > 0

When 1 < x, f ''(x) < 0

So the graph is concave down on (-, -1) U (-1, ) and concave up on (-1, 1). The graph is shown below:

$Graph with vertical asymptotes at x = -1 and x=1, a relative minimum at (0,1), and left and right horizontal asymptotes on the x-axis coming from below.$

PROBLEM 10

Below is the graph of y = f(x) . Sketch the graph of y = f '(x).

$Graph of a function with a vertical asymptote in quadrant 1, then a minimum, and then a maximum in quadrant 1 then a horizontal asymptote.$

Solution

$Graph of the previous f. Also shows f'(c) which has a left and right horizontal asymptote coming from just below the x-axis. It has the same vertical asymptote as f(x), but downwards on the left and upwards on the right. It has a relative maximum just to the right of the x-axis and a relative minimum a bit farther to the right.$

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