College of the Holy Cross MATH 135, section 1 Exam 3 Solutions – Friday, November 22

I. For each of the following functions find the derivative and simplify.

A. (5) f(x) = e^{2x} cos(4x)

Solution: By the product and chain rules,

f′(x) = −4e^{2x} sin(4x) + 2e^{2x} cos(4x) = 2e^{2x}(cos(4x) − 2 sin(4x)).

B. (5) g(x) = \frac{x^{2} + 2}{ln(x)}

Solution: By the quotient rule,

g′(x) = \frac{2x ln(x) − (x^{2} + 2)\frac{1}{x}}{(ln(x))^{2}} = \frac{2x^{2} ln(x) − x^{2} − 2}{x(ln(x))^{2}}.

C. (5) h(x) = tan^{−1}(2x + 1)

Solution: By the derivative rule for inverse tangent and the chain rule,

h′(x) = \frac{2}{1 + (2x + 1)^{2}} .

D. (10) k(x) = x^{sin(x)}

Solution: Since x appears in the base and the exponent, we must use logarithmic differentiation. Writing y = x^{sin(x)}, ln(y) = sin(x) ln(x). So then differentiating implicitly,

\frac{y′}{y} = cos(x) ln(x) + \frac{sin(x)}{x}

and

y′ = x^{sin(x)} (cos(x) ln(x) + \frac{sin(x)}{x}) .

E. (10) Find y′ by implicit differentiation if x^{3}y^{2} + 2y = 3x.

Solution:

3x^{2}y^{2} + x^{3} · 2yy′ + 2y′ = 3,

so

y′ = \frac{3 − 3x^{2}y^{2}}{2x^{3}y + 2} .

II. (20) A rocket is launched vertically and is tracked by a ground station 4 miles from the launch pad. What is the vertical speed of the rocket when its height above the ground is 3 miles and its distance to the ground station is increasing at 3300 miles per hour?

Solution: Let y be the height of the rocket above the launch pad as a function of t and z be the distance to the ground station as a function of t. At each time the position of the rocket, the launch pad, and the ground station form a right triangle, so by the Pythagorean theorem we have

16 + y^{2} = z^{2}.

Differentiating with respect to t:

2y \frac{dy}{dt} = 2z \frac{dz}{dt} .

The vertical speed of the rocket that we want is \frac{dy}{dt}. We know that at the time t when y = 3, z = 5 and \frac{dz}{dt} = 3300. Therefore

\frac{dy}{dt} = \frac{5 · 3300}{3} = 5500m/hr.

III. All parts of this question refer to the function f(x) = −2x^{3} + 24x + 11.

A. (10) Find the critical numbers of f(x).

Solution: We have f′(x) = −6x^{2} + 24 = 0 when x = ±2. f′(x) exists for all real x, so the only critical numbers are x = ±2.

B. (10) What does the Second Derivative Test tell you about the behavior of f at each of these critical numbers?

Solution: f′′(x) = −12x. We see f′′(2) = −24 < 0, so by the Second Derivative Test, f has a local maximum at x = 2. Similarly, f′′(−2) = 24 > 0. Therefore f has a local minimum at x = −2.

IV. All parts of this question refer to the function f(x) = \frac{x}{(2x + 1)^{2}} , for which the first two

derivatives are, in simplified form:

f′(x) = \frac{1 − 2x}{(2x + 1)^{3}}
f′′(x) = \frac{8x − 8}{(2x + 1)^{4}}

A. (5) What is the domain of f(x)?

Solution: Domain of f is all x ≠ −1/2, or (−∞, −1/2) ∪ (−1/2, ∞).

B. (10) Find all critical numbers and determine where f(x) is increasing and decreasing.

Solution: From the formula for f′(x), we see f′(x) = 0 when x = 1/2, and f′(x) exists for all x ≠ −1/2, 1/2. Technical point: x = −1/2 is not a critical number for f, since critical numbers must be in the domain of the function. So the only critical number is x = 1/2. Looking at the sign of f′(x), we see f′(x) < 0 for x in (−∞, −1/2), f′(x) > 0 for x in (−1/2, 1/2), and f′(x) < 0 for x in (1/2, ∞). Therefore f is decreasing on (−∞, −1/2) and (1/2, ∞) and increasing on (−1/2, 1/2).

C. (5) What is the absolute minimum value of f(x) on the interval [0, 3]?

Solution: The critical number 1/2 is in this interval. We have f(1/2) = \frac{1}{8}, f(0) = 0 and f(3) = \frac{3}{49}. The absolute minimum value is f(0) = 0. (Note that by the First Derivative Test, f has a local maximum at x = 1/2, so the minimum must be at one of the endpoints.)

D. (5) Determine the concavity of f(x), and find all inflection points.

Solution: f′′(x) > 0 on (1, ∞) and f′′(x) < 0 on (−∞, −1/2) and (−1/2, 1). The concavity changes at x = 1, so that is the only inflection point.