Mathematics 133 – Intensive Calculus for Science 1 Exam 1 Solutions October 3, 2005

I. First note that f cannot be either linear or exponential because it is increasing for some x and decreasing for other x. A) g(x) is the only linear function, because it is the only one with constant slope. Using

the first two points for instance m = \frac{28−23}{0−(−2)} = \frac{5}{2}. The y-intercept is 28 from the table so g(x) = \frac{5}{2}x + 28.

B) h(x) is exponential. This can be seen because the ratios of successive values is constant:

\frac{18}{2} = \frac{2}{2/9} = \frac{2/9}{2/81} = \frac{2/81}{2/729} = 9.

To find the formula, we try to determine a, b to match the table with h(x) = ba^{x}. From h(0) = 2, we get 2 = ba^{0} = b. Then from h(−2) = 18 we get 18 = 2a^{−2} so 9 = a^{−2}, so a^{2} = 1/9, and a = 1/3. The formula is h(x) = 2 (\frac{1}{3})^{x}.

II. Because the “Explosium” decays exponentially, we know the amount is given by a formula Q(t) = Q_{0}e^{kt} or Q(t) = Q_{0}a^{t}. Either form can be used, and we will show how to do the problem using both of them.

Method 1. Using Q(t) = Q_{0}e^{kt}, the initial amount is Q_{0} = 10, and the half-life is 12 years so

\begin{aligned}5 & = 10e^{k·12} \\ 1/2 & = e^{12k} \\ ln(1/2) & = 12k \\ \frac{ln(1/2)}{12} & = k \end{aligned}

Then we want to determine the time t when Q(t) = 1:

1 = 10e^{\frac{ln(1/2)}{12} t}
1/10 = e^{\frac{ln(1/2)}{12} t}
ln(1/10) = \frac{ln(1/2)}{12}t
\frac{12 ln(1/10)}{ln(1/2)} = t

Method 2. Using Q(t) = Q_{0}a^{t}, we have Q = 10a^{t} and substituting t = 12 (the half-life) we

get 5 = 10a^{12}, so a = (1/2)^{\frac{1}{12}} and Q = 10 ((\frac{1}{2})^{\frac{1}{12}} )^{t} = 10(\frac{1}{2})^{t/12}. Then as before we want to solve for t in the equation 1 = 10 (\frac{1}{2})^{t/12} We get:

1/10 = (\frac{1}{2})^{t/12}
ln(1/10) = (t/12) ln(1/2)
\frac{12 ln(1/10)}{ln(1/2)} = t

(which agrees with the other method, of course!)

III. A) Because this plot is sinusoidal, with a maximum at x = 0, it is simplest to use the form y = A cos(Bx) + C. The amplitude A = (5 − (−3))/2 = 4. The period is 1, so 2π/B = 1 and B = 2π. The graph is also shifted vertically by C = −1:

y = 4 cos(2πx) − 1.

B) This graph looks like a polynomial of degree 3 since there are 2 turning points, a double root at x = 1, and a root at x = 2. This gives y = k(x − 1)^{2}(x − 2) for some constant k. Since the graph “starts high and finishes low” we know k < 0. The exact value can be determined because the graph shows the y-intercept at y = 6:

6 = k(0 − 1)^{2}(0 − 2) = −2k

So k = −3 and the formula is

y = −3(x − 1)^{2}(x − 2).

IV. The graph y = 10^{x+1} is an exponential growth curve (shifted left by 1). So it does pass the horizontal line test (horizontal lines y = c > 0 intersect the graph exactly once). This says there is an inverse function. To find the formula, interchange x and y then solve for y taking logarithms:

x = 10^{y+1}
ln(x) = (y + 1) ln(10)
\frac{ln(x)}{ln(10)} = y + 1

\frac{ln(x)}{ln(10)} − 1 = y

The inverse function is f^{−1}(x) = \frac{ln(x)}{ln(10)} − 1.

V. A) r(x) has vertical asymptotes where the bottom is zero: 16 − x^{2} = 0, so x = ±4. B) r(x) has a horizontal asymptote at y = −2 because as x → ±∞,

\frac{2x^{2} − x}{16 − x^{2}} = \frac{2 − \frac{1}{x}}{\frac{16}{x^{2}} − 1} → −2

C) The graph y = r(x) crosses the x-axis where the top is zero: 0 = 2x^{2} − x = x(2x − 1), so x = 0 or x = 1/2.