Mathematics 133 – Intensive Calculus for Science 1 Sample Exam II – October 26, 2001

I. Give a formula for a rational function r(x) = \frac{ax^{2}+b}{x^{2}+dx+e} which has vertical asymptotes at x = 1, 3, x-axis intercepts at x = −2 and x = 2, and which has y = 6 as a horizontal asymptote.

Solution: The numerator must have zeroes at x = ±2, so it looks like ax^{2} + b = a(x + 2)(x − 2) = ax^{2} − 4a. The denominator must have zeroes at x = 1, 3, so it looks like x^{2} + dx + e = (x − 1)(x − 3) = x^{2} − 4x + 3. The y-level of the horizontal asymptote is the limit of r(x) as x → ±∞, which is

lim_{x→±∞} \frac{ax^{2} − 4a}{x^{2} − 4x + 3} = lim_{x→±∞} \frac{ax^{2} − 4a}{x^{2} − 4x + 3} · \frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}}

= lim_{x→±∞} \frac{a − 4\frac{a}{x^{2}}}{1 − \frac{4}{x} + \frac{3}{x^{2}}}

= a

Hence a = 6, and the full equation is

r(x) = \frac{6x^{2} − 24}{x^{2} − 4x + 3}

II. The following graph shows y = f(x). Using the information here, construct a “qualitative” plot of y = f′(x).

III. Using the limit definition, find f′(x) for f(x) = x^{2} − x + 4.

Solution:

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

= lim_{h→0} \frac{(x + h)^{2} − (x + h) + 4 − x^{2} + x − 4}{h}

= lim_{h→0} \frac{x^{2} + 2xh + h^{2} − x − h + 4 − x^{2} + x − 4}{h}

= lim_{h→0} \frac{2xh − h + h^{2}}{h}

= lim_{h→0} 2x − 1 + h

= 2x − 1

IV. Find derivatives of each of the following functions by applying the appropriate “shortcut” derivative rules:

A) (10) f(x) = 3^{x} + x^{6} − \frac{3}{\sqrt[3]{x}}

Solution: f(x) = 3^{x} + x^{6} − 3x^{−1/3}, so applying the sum, power, and exponential rules:

f′(x) = 3^{x} ln(3) + 6x^{5} + x^{−4/3}

B) (10) g(x) = x^{e^{2}} + (e^{x})^{2}

Solution: The first term is a power; the second can be rewritten as the exponential e^{2x} = (e^{2})^{x}. So

g′(x) = e^{2}x^{e^{2}−1} + (e^{2})^{x} ln(e^{2}) = e^{2}x^{e^{2}−1} + 2e^{2x}

C) (10) h(x) = \frac{4x^{2}−3}{x^{5}}

Solution: This is one where you need “preliminary algebra” (if you don’t use the quotient rule!):

h(x) = 4x^{−3} − 3x^{−5}  so
h′(x) = −12x^{−4} + 15x^{−6}

= \frac{−12x^{2} + 15}{x^{6}}

V. The price of an antique rocking chair at time t years after 1975 is approximated by P (t) = 100(1.25)^{t}. What was the price in 1990, and how fast was the price changing then? Was it increasing or decreasing?

Solution: 1990 is 15 years after 1975, so P (15) = 100(1.25)^{15} ≐ 2842.17 and P′(t) = 100(1.25)^{t} ln(1.25), so P′(15) = 100(1.25)^{15} ln(1.25) ≐ 624.21. The price was increasing (P′(t) > 0) at a rate of 624.12 $/year.

VI. It costs C(q) = 50q^{1/4}+1000 dollars for an independent record company to manufacture q copies of its star alternative band’s latest CD. Compute C′′(q). Is the graph of the cost function concave up or down? What does your answer mean in everyday terms?

Solution: C′(q) = \frac{25}{2}q^{−3/4}, so C′′(q) = \frac{−75}{8}q^{−7/4}. This shows C′′(q) is always negative, so the graph is concave down. In everyday terms, this says that the more CD’s are produced, the lower the cost per CD (in economics, this is called an “economy of scale”).