Mathematics 133 – Intensive Calculus for Science 1 Solutions – Final Examination December 11, 2001

I. A) The function plotted looks most like a sinusoidal oscillation with amplitude A = 5,

vertical shift C = 2, and period 8, so B = 2π/8 = π/4. One possible formula is

y = 5 sin (\frac{πx}{4}) + 2

Note: It could also be argued that this looks like a polynomial function, but then there would have be at least four roots and four turning points, so the degree would have to be 5 at least.

B) There is no inverse function for f on 0 ≤ x ≤ 4 since that part of the graph does not pass the horizontal line test.

II. There are x-axis intercepts at x = −2 and x = 3, and vertical asymptotes at x = −1, 1. So the rational function looks like:

r(x) = \frac{c(x + 2)(x − 3)}{(x − 1)(x + 1)} = \frac{c(x^{2} − x − 6)}{x^{2} − 1}

for some constant c. To determine the value of c, we can look at the y-intercept. When x = 0, y = −6, so

−6 = \frac{c(0^{2} − 0 − 6)}{0^{2} − 1} = \frac{−6c}{−1} = 6c.

So c = −1, and

r(x) = \frac{−(x + 2)(x − 3)}{(x − 1)(x + 1)} = \frac{−x^{2} + x + 6}{x^{2} − 1}

III. The function f is approximately linear since the given x values are equally spaced, and all the slopes are .21/.2 = 1.05. For instance, between the first two points

m = \frac{Δy}{Δx} = \frac{2.51 − 2.30}{.4 − .2} = \frac{.21}{.2} = 1.05

The equation of the line is y = 1.05x + b for some b. Substituting x = .2, y = 2.30, we have 2.30 = 1.05 · .2 + b, so b = 2.09. The equation is y = 1.05x + 2.09.

IV. A) The temperature at time t = 0 is T (0) = 70 + 80e^{−0/3} = 70 + 80 = 150 degrees F. B) The rate of change of the temperature at t = 10 is T′(10) (the derivative). We have

T′(t) = 80e^{−t/3}(−1/3) (by the chain rule). So T′(10) = −80e^{−10/3}/3 ≐ −.951 degrees F per minute (must be negative since the chocolate is cooling off and tending to the room temperature over time).

C) We must solve the equation 100 = 70 + 80e^{−t/3} for t. Subtract 70 from both sides to get 30 = 80e^{−t/3}. Then divide by 80 to get 3/8 = e^{−t/3}. Take natural logs of both sides: ln(3/8) = −t/3, so t = −3 ln(3/8) ≐ 2.94 minutes.

V. A) If f(x) = \frac{1}{x+1},

f′(x) = 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 + h + 1)(x + 1)}

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

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

= \frac{−1}{(x + 1)^{2}}

B) The function g(x) is g(x) = 3x^{4} + 3x^{−1/2} + 2x^{1/3}. So

g′(x) = 12x^{3} − \frac{3}{2}x^{−3/2} + \frac{2}{3}x^{−2/3}

C) Using the quotient and chain rules:

\frac{d}{dx} (\frac{tan(x) + x}{cos(2x)}) = \frac{cos(2x)\frac{d}{dx}(tan(x) + x) − (tan(x) + x)\frac{d}{dx}(cos(2x))}{(cos(2x))^{2}}

= \frac{cos(2x)(sec^{2}(x) + 1) − (tan(x) + x)(− sin(2x) · 2)}{cos^{2}(2x)}

= \frac{cos(2x)(sec^{2}(x) + 1) + 2(tan(x) + x) sin(2x)}{cos^{2}(2x)}

D) By the chain rule:

\frac{d}{dx}(ln(x^{2} + 3^{x})) = \frac{1}{x^{2} + 3^{x}} (2x + 3^{x} ln(3))

E) By the chain rule again:

\frac{d}{dx}(arcsin(12x + 2)) = \frac{1}{\sqrt{1 − (12x + 2)^{2}}} · 12

VI. Qualitatively,

VII. The given graph shows y = f′(x), not y = f(x).

A) f is increasing on intervals where f′(x) > 0 (graph of f′ above x-axis): approximately

−1.3 < x < −.5, 0 < x < .5, and 1.3 < x < 1.5. B) The critical points are approximately x = −1.3, −.5, 0, .5, and 1.3. By the first deriva-

tive test, −1.3 is a local minimum, −.5 is a local maximum, 0 is a local minimum, .5 is a local maximum, and 1.3 is another local minimum of f.

C) The graph of f is concave up on intervals where f′ is increasing, so approximately

−1.5 < x < −1, −.3 < x < .3, and 1 < x < 1.5. D) The graph of f has inflection points where the concavity changes, or where f′ changes

from increasing to decreasing or vice versa. These are at approximately x = −1, −.3, .3, 1 (the critical points of f′).

VIII. Here is a picture of the window as described, with dimensions labeled:

Note that the radius of the semicircle is x/2, one half of that side of the rectangle. The perimeter is the sum of the lengths of three sides of the rectangle, plus the length of the semicircle:

30 = x + 2y + \frac{1}{2} · 2π \frac{x}{2} = (1 + \frac{π}{2} ) x + 2y

From this we can solve for y in terms of x:

(1)  y = \frac{30 − (1 + \frac{π}{2} ) x}{2} = 15 − (\frac{1}{2} + \frac{π}{4} ) x

The area of the window is the sum of the areas of the rectangle and the semicircle:

A = xy + \frac{1}{2} · π (\frac{x}{2})^{2}

Substituting for y from (1) and simplifying:

A = x (15 − (\frac{1}{2} + \frac{π}{4} ) x) + \frac{πx^{2}}{8} = 15x − (\frac{1}{2} + \frac{π}{8} ) x^{2}

We are now ready to maximize the area. First, we find critical points:

A′(x) = 15 − (1 + \frac{π}{4} ) x = 0

has only one solution:

x = \frac{15}{1 + \frac{π}{4}} ≐ 8.4

Then from (1)

y = 15 − (\frac{1}{2} + \frac{π}{4} ) · 8.4 ≐ 4.2

This is a local maximum since by the second derivative test,

A′′(x) = − (1 + \frac{π}{4} ) < 0

IX. This is a true statement. The vertical stretching by a factor of 2 gives y = 2 · 2^{−x} = 2^{−x+1} = 2^{−(x−1)}, which is the horizontal shift of y = 2^{−x} by one unit to the right.