Mathematics 133–Intensive Calculus for Science 1 Solutions for Exam 2 November 4, 2005

I. A) From the graph, f(x) appears to be differentiable everywhere except x = 0, where

there is a sharp corner. B) The derivative y = f′(x):

Note: f′(x) > 0 is positive on intervals where f(x) is increasing and negative on intervals where f(x) is decreasing. At x = 0, there is a jump discontinuity on the graph y = f′(x) because of the sharp corner on the graph y = f(x).

II.

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^{2} − h}{h}

= lim_{h→0} 2x + h − 1

= 2x − 1

III. A) (Product, exponential, and power rules)

f′(x) = 2^{x} ln(2)(x^{4}+3x^{−1/3})+2^{x}(4x^{3}−x^{−4/3}) = 2^{x}(ln(2)x^{4}+3 ln(2)x^{−1/3}+4x^{3}−x^{−4/3})

B) By the chain rule (twice):

g′(x) = 3(e^{πx} + 2)^{2}πe^{πx}

C) By the quotient rule:

h′(x) = \frac{(2x^{2} + 3x)(2x) − (x^{2} − 3)(4x + 3)}{(2x^{2} + 3)^{2}} = \frac{3x^{2} + 12x + 9}{(2x^{2} + 3)^{2}}

IV. The price in 2000 was P (50) = 100(1.05)^{5}0 ≐ $1146.74. The price was changing at P′(50) = 100(1.05)^{5}0 ln(1.05) ≐ 55.95 dollars per year. It was increasing because P′(50) > 0.

V. C′(q) = \frac{50}{4}q^{−3/4}, so C′′(q) = \frac{−150}{16}q^{−7/4}. The cost function graph is concave down for q > 0 since C′′(q) < 0. In everyday terms, this means that as the number of copies is increased, the cost per copy decreases.