Mathematics 133 – Intensive Calculus for Science 1 Solutions – Exam 3 December 3, 2001

I. A) Using the derivative rule for ln and the chain rule:

f′(x) = \frac{1}{x} + 2^{4x} ln(2) · 4 = \frac{1}{x} + 4 ln(2)2^{4x}

B) Using the product and chain rules and the derivative rules for sin and cos:

g′(x) = sin(x) · − sin(x^{2}) · 2x + cos(x^{2}) · cos(x) = −2x sin(x) sin(x^{2}) + cos(x) cos(x^{2})

C) Using the derivative rule for arctan and the chain rule:

h′(x) = \frac{1}{1 + (x^{2} + e^{2x} + 1)^{2}} · (2x + 2e^{2x})

II. A) Using the quotient rule,

f′(x) = \frac{(x^{2} + 4) · 1 − x · 2x}{(x^{2} + 4)^{2}}

= \frac{−x^{2} + 4}{(x^{2} + 4)^{2}}

= \frac{(2 − x)(2 + x)}{(x^{2} + 4)^{2}}

(factoring using the pattern for a difference of squares). B) From the factored form of the numerator, the critical points (x where f′(x) = 0) are

x = ±2. The denominator of f′(x) is always positive, so the sign of f′ is determined by the numerator. −x^{2} + 4 is > 0 if −2 < x < 2 and positive when x > 2 or x < −2. The First Derivative Test shows that x = −2 is a local minimum, while x = 2 is a local maximum.

C) Differentiating f′ with the quotient rule again, we get

f′′(x) = \frac{(x^{2} + 4)^{2} · (−2x) − (−x^{2} + 4) · 2(x^{2} + 4) · 2x}{(x^{2} + 4)^{4}}

We want to simplify this. Note that there are several common factors in the terms on the top of this expression. WE DO NOT WANT TO MULTIPLY THIS OUT!! Instead, we factor out 2x(x^{2}+4), cancel with the denominator, and simplify to obtain:

\begin{aligned}f′′(x) & = \frac{2x(x^{2} + 4)(−(x^{2} + 4) − 2(−x^{2} + 4))}{(x^{2} + 4)^{4}} \\ & = \frac{2x(x^{2} − 12)}{(x^{2} + 4)^{2}} \end{aligned}

D) The inflection points of this function are at the roots of the numerator of f′′(x) – x = 0, ±\sqrt{12}. The second derivative changes sign at each of these.

III. Let x, y be the overall dimensions of the poster. The printed region is x − 4 inches by y − 3 inches because of the margins. Then we have xy = 400, and we want to maximize the area of the printed region, which is A = (x − 4)(y − 3). Solving for y from the equation xy = 400, y = 400/x, so

A(x) = (x − 4) (\frac{400}{x} − 3)

= 400 − \frac{1600}{x} − 3x + 12

= 412 − \frac{1600}{x} − 3x

A′(x) = \frac{1600}{x^{2}} − 3

The critical points of the area function are at the solutions of A′(x) = 0, so

\frac{1600}{x^{2}} − 3 = 0
\frac{1600}{x^{2}} = 3
x^{2} = \frac{1600}{3}
x = \sqrt{\frac{1600}{3}}
≐ 23.1

Then y = \frac{400}{23.1} ≐ 17.3. This is a maximum for the area because A′′(x) = −\frac{3200}{x^{3}}, so

A′′ (\sqrt{\frac{1600}{3}}) < 0 (Second Derivative Test says that if f′′ is negative at the critical point, then f has a local maximum there).

IV. A) True: Say p_{1} < p_{2}. If we assume that f′ is continuous for all x, then f′ must be zero

somewhere between p_{1} and p_{2}. If not, f′ would have to have the same sign everywhere between p_{1} and p_{2}. That sign would have to be negative because p_{1} is a local max. But f′(x) cannot be negative to the left of p_{2} if p_{2} also a local max.

B) True: By implicit differentiation, 2y \frac{dy}{dx} + x\frac{dy}{dx} + y = 0, so

\frac{dy}{dx} = \frac{−y}{2y + x}

In this case we can solve explicitly for y as a function of x with the quadratic formula:

y = \frac{−x ± \sqrt{x^{2} + 4}}{2}

Each of these functions is defined for all x, and 2y + x = \sqrt{x^{2} + 4} ≠ 0 for all x. Hence \frac{dy}{dx} exists.