College of the Holy Cross, Fall Semester, 2007 MATH 131, Section 01, Final Exam Solutions December 10

1. [5 points each] Circle the number of the graph showing each of the following functions. Note that there is one more graph than there are formulas.

Solution:

(a) f(x) = 2 − 2 sin(πx) is graph III.

(b) f(x) = \frac{x^{2}}{x^{2} − 1} is graph V. (Note the vertical asymptotes at x = ±1.)

(c) f(x) = ln(\sqrt{x}) is graph I.

(d) f(x) = e^{x} +e^{−x} is graph II. (This is not graph IV which has a somewhat similar shape, since f(0) = e^{0} + e^{0} = 2. This matches II but not IV.)

I: II:

III: IV:

2.

(a) [5 points] What is the domain of the function f(x) = \sqrt{x^{2} − 4}?

Solution: The domain is the set of all real x with x^{2}−4 ≥ 0, or x in (−∞, 2]∪[2, +∞).

(b) [10 points] Find a formula for the inverse function of f(x) = \frac{x}{3} − 1 and plot y = f(x) and y = f^{−1}(x) on the same coordinate axes below.

Solution: To find the formula of f^{−1}, set y = \frac{x}{3} − 1 and solve for x: x = 3(y + 1) = 3y + 3. Switch variables to get y = f^{−1}(x) = 3x + 3. The graph y = f(x) is a straight line with slope \frac{1}{3} and y-intercept at −1. The graph y = f^{−1}(x) is a straight line with slope 3 and y-intercept at y = 3:

(c) [5 points] Find the interval of all real x satisfying −π/4 < tan^{−1}(x) < π/3.

Solution: Since the tangent function is increasing, this set of inequalities is the same as tan(−π/4) < x < tan(π/3), or −1 < x < \sqrt{3}.

(d) [5 points] Express as a single logarithm: ln(6) − ln(12) + 5 ln(2).

Solution: Using properties of logarithms, this equals ln (\frac{6 · 2^{5}}{12} ) = ln(16).

3. Compute the following limits [5 points each]. Any legal method is OK.

(a) lim_{x→3^{−}} \frac{|x − 3|}{x^{2} − 2x − 3}

MATH 131-01, Final Exam 3 Solution: The denominator factors as x^{2} −2x−3 = (x−3)(x+1). But \frac{|x − 3|}{x − 3} = −1 when x < 3 (which is what we want for the left-hand limit). Hence

lim_{x→3^{−}} \frac{|x − 3|}{x^{2} − 2x − 3} = lim_{x→3^{−1}} \frac{−1}{x + 1} = \frac{−1}{4}.

(b) lim_{x→−1} \frac{x^{3} − x}{x^{2} + 5x + 4}

Solution: This one is indeterminate of the form 0/0 and can be done via “preliminary algebra” or L’Hopital’s Rule. If we use L’Hopital,

lim_{x→−1} \frac{x^{3} − x}{x^{2} + 5x + 4} = lim_{x→−1} \frac{3x^{2} − 1}{2x + 5} = \frac{2}{3}.

(c) lim_{x→+∞} \frac{ln(x^{2})}{x^{3/2}}

Solution: Indeterminate of the form ∞/∞, so we use L’Hopital, then simplify:

lim_{x→+∞} \frac{ln(x^{2})}{x^{3/2}} = lim_{x→+∞} \frac{\frac{2x}{x^{2}}}{(3/2)x^{1/2}} = lim_{x→+∞} \frac{4}{3x^{3/2}} = 0.

4.

(a) [5 points] State the limit definition of the derivative.

Solution: The derivative is

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h}

provided that the limit exists.

(b) [10 points] Use the definition to compute f′(x) for f(x) = \frac{1}{x − 1}.

Solution: We compute

lim_{h→0} \frac{f(x + h) − f(x)}{h} = lim_{h→0} \frac{\frac{1}{x+h−1} − \frac{1}{x−1}}{h}
= lim_{h→0} \frac{(x − 1) − (x + h − 1)}{h(x − 1)(x + h − 1)}
= lim_{h→0} \frac{−h}{h(x − 1)(x + h − 1)}
= lim_{h→0} \frac{−1}{(x − 1)(x + h − 1)}
= \frac{−1}{(x − 1)^{2}}

(c) [10 points] Find the equation of the tangent line to the graph y = \frac{1}{x−1} at the point (5, 1/4).

Solution: The tangent line has slope f′(5) = \frac{−1}{16} and passes through the point (5, 1/4) on the graph of f. So the equation is

y − 1/4 = \frac{−1}{16}(x − 5).

5. The following graph shows y = g′(x), the derivative of some function g(x).

Note: this is y = g′(x), not y = g(x).

(a) [5 points] At which of the points x = 0, a, b, c, d, e, f does the function g(x) have local maxima?

Solution: By the first derivative test, g(x) has local maxima where g′(x) changes from positive to negative: at x = c, e.

(b) [5 points] At which of the points x = 0, a, b, c, d, e, f does the function g(x) have local minima?

Solution: By the first derivative test, g(x) has local minima where g′(x) changes from negative to positive: at x = a, d, f.

(c) [5 points] On the interval (a, b), is y = g(x) concave up or down?

Solution: On the interval (a, b), g′(x) is increasing, so (g′)′(x) = g′′(x) > 0. This implies y = g(x) is concave up.

6. Compute the following derivatives using the derivative rules. You need not simplify. [5 points each]

(a) f(x) = 3x^{5} + \frac{5}{x^{3}} + e^{x}.

Solution: f′(x) = 15x^{4} − 15x^{−4} + e^{x}.

(b) g(x) = \frac{sin(x) + 1}{sin^{−1}(x) + 1}.

Solution: By the quotient rule,

g′(x) = \frac{(sin^{−1}(x) + 1) cos(x) − (sin(x) + 1)\frac{1}{\sqrt{1−x^{2}}}}{(sin^{−1}(x) + 1)^{2}} .

(c) h(x) = \sqrt{ln(x) + 4 cos(tan(x))}.

Solution: By the chain rule and the derivative rules for ln, cos, tan:

h′(x) = \frac{1}{2} (ln(x) + 4 cos(tan(x)))^{−1/2} · (\frac{1}{x} − 4 sin(tan(x)) sec^{2}(x)) .

(d) j(x) = (x^{2} + 3)^{(x+2)}.

Solution: Using logarithmic differentiation, ln(j(x)) = (x + 2) ln(x^{2} + 3). So then by the product rule

\frac{j′(x)}{j(x)} = \frac{2x(x + 2)}{x^{2} + 2} + ln(x^{2} + 3)

Hence

j′(x) = (x^{2} + 3)^{(x+2)} (\frac{2x(x + 2)}{x^{2} + 2} + ln(x^{2} + 3)) .

(e) Find \frac{dy}{dx} if x^{3} + y^{3} = 18xy.

Solution: Differentiating implicitly,

3x^{2} + 3y^{2}y′ = 18y + 18xy′.

Then we solve algebraically for y′ = \frac{dy}{dx}:

y′ = \frac{18y − 3x^{2}}{3y^{2} − 18x} = \frac{6y − x^{2}}{y^{2} − 6x}.

7. All parts of this question refer to the function

f(x) = \frac{(x + 1)^{2}}{x^{2} + 1} .

(a) [5 points] Compute f′(x) and simplify.

Solution: By the quotient and chain rules:

f′(x) = \frac{(x^{2} + 1) · 2(x + 1) − (x + 1)^{2} · 2x}{(x^{2} + 1)^{2}} = \frac{−2(x^{2} − 1)}{(x^{2} + 1)^{2}}.

(b) [10 points] Find all critical numbers of f and determine the intervals on which f is increasing and decreasing.

Solution: f′(x) = 0 at x = ±1. These are the critical numbers. f′(x) > 0 so f(x) is increasing on (−1, 1). f′(x) < 0 so f(x) is decreasing on (−∞, −1) ∪ (1, +∞).

(c) [5 points] Given: the second derivative in simplified form is

f′′(x) = \frac{4x(x^{2} − 3)}{(x^{2} + 1)^{3}}.

Determine all points of inflection and the intervals where y = f(x) is concave up and concave down.

Solution: From the form of the numerator, f′′(x) = 0 at x = 0, ±\sqrt{3}. f′′(x) > 0 so f(x) is concave up on (−\sqrt{3}, 0) ∪ (\sqrt{3}, +∞). f′′(x) < 0 so f(x) is concave down on (−∞, −\sqrt{3})∪(0, \sqrt{3}). The concavity changes at −\sqrt{3}, 0, \sqrt{3}, so all three are inflections.

(d) [5 points] Find the absolute maximum and minimum of f(x) on the interval [0, 5].

Solution: We compare f(0) = 1, f(1) = 2 (the value at the critical number in the interval), and 1 < f(5) = 18/13 < 2. Hence the absolute minimum on this interval is f(0) = 1, and the absolute maximum is f(1) = 2.

8. [20 points] A potter forms 20 cm^{3} of clay into a cylindrical tube. As she rolls the tube on a flat surface, its length L increases and its radius r decreases but the quantity of clay stays constant. If the length is increasing at the rate of .1 cm/sec, find the rate at which the radius is changing when the radius is 1cm. (Note: the volume of a cylinder of radius r and height h is V = πr^{2}h.)

Solution: The volume of the clay is 20 = πr^{2}L at all times. Taking derivatives with respect to time, and using the product rule,

0 = 2πrL\frac{dr}{dt} + πr^{2} \frac{dL}{dt}.

When r = 1, \frac{dL}{dt} = +0.1 and from the volume equation L = \frac{20}{π}. Hence the radius is changing at the rate:

\frac{dr}{dt} = \frac{−πr^{2} \frac{dL}{dt}}{2πrL} = \frac{−π(1)^{2}(0.1)}{2π(1)\frac{20}{π}} = \frac{−π}{400} ≐ −0.0079 cm/sec.

9. [20 points] A rectangular swimming pool with sides x, y is to be built with an area of 1800 square feet. The owner wants 5-foot wide decks along either side and 10-foot wide decks at the two ends. Find the dimensions of the rectangular lot of smallest area on which such a swimming pool can be built. Be sure you say how you know your answer gives a minimum of the area of the lot.

Solution: The area of the pool is xy = 1800, so y = 1800/x. If we let x be width of the pool, then the width of pool plus side decks is x + 10. The length of the pool plus the end decks is then y + 20. The total area of the rectangle containing the pool and the decks is

A = (x + 10)(y + 20) = xy + 10y + 20x + 200 = 2000 + \frac{18000}{x} + 20x.

Taking derivative with respect to x, and setting equal zero to find critical numbers, we find

0 = A′(x) = 20 − \frac{18000}{x^{2}}

so x = ±\sqrt{18000/20} = ±\sqrt{900} = ±30. The negative root is irrelevant here. So x = 30 is the only critical number we need to consider. By the equation y = 1800/x we get y = 60. Now, at x = 30, A′′(x) = \frac{36000}{30^{3}} > 0. Hence this gives a minimum of the area of the pool plus the decks. The final answer is that the lot should be x + 10 = 40 by y + 20 = 80 feet to fit the pool and the decks.

10. On and near the surface of the moon, the acceleration of gravity is −1.66 meters per second per second.

(a) [5 points] An astronaut throws a moon rock upward from a height of 1.5 meters above the surface with an initial velocity of 3 meters per second. Find the height of the rock above the surface as a function of time t in seconds.

Solution: We antidifferentiate once to find the velocity:

v(t) = −1.66t + c.

To get v(0) = 3, c = 3. Then we antidifferentiate again to get the height:

h(t) = −.83t^{2} + 3t + d.

To get h(0) = 1.5, d = 1.5. So h(t) = −.83t^{2} + 3t + 1.5.

(b) [3 points] How long does it take for the rock from part a to fall back to the surface of the moon?

Solution: We need to solve the equation h(t) = −.83t^{2} + 3t + 1.5 = 0. The only way to do this is via the quadratic formula:

t = \frac{−3 ± \sqrt{9 − 4(−.83)(1.5)}}{−1.66}.

The positive root is about t = 4.06 sec.

(c) [2 points] What is the velocity of the rock from part a when it strikes the surface of the moon?

Solution: Substitute t = 4.06 into the formula for v(t) above: v(4.06) ≐ −3.74 m/sec.