Mathematics 136 – Advanced Placement Calculus Solutions to Ungraded Problems on Problem Set 1 September 15, 2009

1.2/15. (a) We want T as a function of N. The slope of the linear function is given by \frac{80−70}{173−113} = \frac{10}{60} = \frac{1}{6}. By the point-slope form, T − 70 = \frac{1}{6}(N − 113), so T = \frac{1}{6}N + \frac{307}{6}.

(b) The slope is \frac{1}{6}. The meaning of this is that each increase of 6 chirps per minute corresponds to a 1 degree increase in temperature.

(c) If N = 150, then T = \frac{1}{6}(150) + \frac{307}{6} ≐ 76 degrees Fahrenheit.

1.3/7. The graph y = \sqrt{3x − x^{2}} is shifted 4 units to the right, reflected over the x-axis and shifted 1 unit down:

y = −\sqrt{3(x + 4) − (x + 4)^{2}} − 1 = −\sqrt{−x^{2} − 5x − 4} − 1.

1.3/16. The graph y = \frac{1}{x−4} is obtained by shifting the graph y = \frac{1}{x} four units horizontally to the right. (This also shifts the vertical asymptote to the line x = 4.)

1.3/24. The graph y = | cos(πx)| is obtained from y = cos(x) by compressing horizontally (the period becomes 2 rather than 2π), and reflecting the parts of the graph y = cos(πx) that lie below the x-axis over the x-axis. See Figure 1 (on the back page).

1.5/21. Since (1, 6) and (3, 24) must be on the graph y = Ca^{x}, we get the two equations

6 = Ca^{1} = Ca
24 = Ca^{3}.

Solving for C in the first equation, C = \frac{6}{a}. So substituting into the second equation, we get 24 = \frac{6}{a}a^{3} = 6a^{2}. Since a > 0, this shows a = 2, and hence C = 3. The equation is y = 3 · 2^{x}.

1.6/26. First we solve the equation y = \frac{e^{x}}{1+2e^{x}} for x as a function of y:

(1 + 2e^{x})y = e^{x} (after clearing denominators)
y + 2ye^{x} = e^{x}
y = (1 − 2y)e^{x}
\frac{y}{1 − 2y} = e^{x}
ln (\frac{y}{1 − 2y} ) = x (after taking logarithms to solve for x) .

Then f^{−1}(x) = ln (\frac{y}{1−2y} ) is the inverse function (swap x and y).

Figure 1: Plot for 1.3/24

1.6/51. Solve for x in the exponent by taking logarithms. Using natural log,

ln (2^{x−5}) = ln(3)
(x − 5) ln(2) = ln(3)
ln(2)x = ln(3) + 5 ln(2)
x = \frac{ln(3) + 5 ln(2)}{ln(2)} = \frac{ln(96)}{ln(2)}, or 5 + \frac{ln(3)}{ln(2)}.

2.2/5 (a) lim_{t→0^{−}} g(t) = −1.

(b) lim_{t→0^{+}} g(t) = −2.

(c) lim_{t→0} g(t) does not exist since the one-sided limits are different.

(d) lim_{t→2^{−}} g(t) = 2.

(e) lim_{t→2^{+}} g(t) = 0.

(f) lim_{t→2} g(t) does not exist (same reason as in (c)).

(g) g(2) = 1

(h) lim_{t→4} g(t) = 3.

2.2/15. There are infinitely many correct graphs here. They would all have jump discontinuities at x = 3, and removable discontinuities at x = −2.