QUIZ #7: CALCULUS 1A (Stankova) Wednesday, March 10, 2004 Section 10:00–11:00 (Voight)

Problem 1. Evaluate

lim_{x→e} \frac{e^{ln x} − e}{x − e}.

Explain your work.

Solution. We match to the template

f′(a) = lim_{x→a} \frac{f(x) − f(a)}{x − a}.

We take f(x) = e^{ln x} and a = e, and verify that f(a) = e^{ln e} = e indeed. Therefore the limit is equal to f′(e). We compute

f′(x) = \frac{1}{x}e^{ln x}

so f′(e) = e/e = 1, which is equal to the limit.

Or, you can simply note that e^{ln x} = x, so we just have

lim_{x→e} \frac{x − e}{x − e} = 1.

Problem 2. Evaluate

lim_{x→0} \frac{sin((3 + x)^{2}) − sin 9}{x}.

Explain your work.

Solution. We again match to the template, and let f(x) = sin((3+x)^{2}) and a = 0. We verify that f(0) = sin 9, so the limit is equal to f′(0). We compute by the chain rule:

f′(x) = cos((3 + x)^{2})((3 + x)^{2})′ = cos((3 + x)^{2})(2(3 + x))

and so f′(0) = 6 cos 9, the value of the limit.

QUIZ #7: CALCULUS 1A (Stankova) Wednesday, March 10, 2004 Section 11:00–12:00 (Voight)

Problem 1. Sketch the graph of f by hand and use your sketch to find the absolute (global) and local maximum and minimum values of f.

f(x) = \begin{cases}2x^{2} − 1, & if − 1 ≤ x < 0; \\ 1 − (x − 1)^{2}, & if 0 ≤ x ≤ 2. \end{cases}

Solution. We have the graph:

Looking at the graph, we see that f has an absolute maximum at x = −1 and x = 1. It has no local or absolute minimum, because if it did it would be at x = 0 but f jumps at that point. It has a local maximum at x = 1.