Math 131, Midterm 2 Solutions 1

College of the Holy Cross, Fall Semester, 2004 Math 131, Midterm 2 Solutions

1. [5 points each] Compute the following derivatives. You may use any correct method.

(a) \frac{d}{dx}(5x\sqrt{x} − \frac{2}{x^{3}} + 11x − 4)

Solution First rewrite x\sqrt{x} = x^{3/2} and \frac{2}{x^{3}} = 2x^{−3}. Then by the power rule,

d_{\overline{dx(5x^{3/2}−2x^{−3}+11x−4)}} =15_{\overline{2x^{1/2}+6x^{−4}+11.}}

(b) \frac{d}{dt}(t^{2}e^{−5t})

Solution By the product rule,

d_{\overline{dt(t^{2}e^{−5t})=t^{2} \frac{d}{dt} (e^{−5t})+e^{−5t} \frac{d}{dt} (t^{2})}}

Using the chain rule on the first term, this equals

t^{2}e^{−5t}(−5) + e^{−5t}(2t)

which simplifies to (2t − 5t^{2})e^{−5t}.

(c) \frac{d}{dz} 8(z^{2} + 4 cos z + 2)^{3}

Math 131, Midterm 2 Solutions 2 Solution By the chain rule,

d_{\overline{dz8(z^{2}+4 cos z+2)^{3}=24(z^{2}+4 cos z+2)^{2}(2z−4 sin z).}}

(d) \frac{d}{dx}(\frac{x}{sin x})

Solution By the quotient rule,

d_{\overline{dx(\frac{x}{sin x})}}

=sin x − x cos x_{\overline{sin^{2} x}}

2. [15 points] The graph of a function f is shown below with several points marked. Check the appropriate boxes.

A  B  C D E  F  G

Point at which A B C D E F G f is not continuous

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✓

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f is not differentiable

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3. [5 points each] Compute the indicated limits. Show all work for full credit.

(a) lim_{x→1} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution Since lim_{x→1} x^{2} − 4x + 4 = 1 ≠ 0,

lim_{x→1} \frac{3x^{2}−5x−2}{x^{2}−4x+4}

=lim_{x→1} 3x^{2} − 5x − 2\overline{lim_{x→1} x^{2} − 4x + 4}

=-4_{\overline{1=−4}}

(b) lim_{x→2} \frac{3x^{2} − 5x − 2}{x^{2} − 4x + 4}

Solution Since lim_{x→2} 3x^{2} − 5x − 2 = 0 and

lim_{x→2} x^{2} − 4x + 4 = 0, we cannot just “plug in” to compute the limit. Factoring gives

3x^{2} − 5x − 2_{\overline{x^{2}−4x+4= \frac{(x−2)(3x+1)}{(x−2)^{2}}}}

=3x+1_{\overline{x−2 for x≠2}}

The resulting function \frac{3x+1}{x−2} has a vertical asymptote at x = 2 since x = 2 is a root of the denominator and not a root of the numerator. Consequently

lim_{x→2} \frac{3x^{2}−5x−2}{x^{2}−4x+4} = lim_{x→2} \frac{3x+1}{x−2}

does not exist.

(c) lim_{x→1^{−}} 3 · \frac{x − 1}{|x − 1|} + 1

Solution For x < 1, x − 1 < 0 so |x − 1| = −(x − 1) and therefore

3·\frac{x−1}{|x−1|} + 1 = 3 · \frac{x−1}{−(x−1)} + 1 = −3 + 1 = −2

so

lim_{x→1^{−}} 3 · \frac{x−1}{|x−1|} + 1 = −2.

4. [5 points each] Each part refers to the graph shown.

y = f(x) A  B  C  D  E  F

(a) Find all intervals on which f(x) > 0. Solution The graph lies above the x-axis between B and E, so f(x) > 0 for B < x < E.