QUIZ #8: CALCULUS 1A (Stankova) Wednesday, March 17, 2004 Section 10:00–11:00 (Voight)

Problem 1. Verify that the function

f(x) = x^{3} + 2x − 2

satisfies the hypotheses of the Mean Value Theorem on the interval [0, 1].

Then find all numbers c that satisfy the conclusion of the Mean Value Theorem.

Solution. The function f is continuous on [0, 1] and differentiable on (0, 1) because it is a polynomial, therefore it satisfies the MVT.

We compute that

f′(x) = 3x^{2} + 2

and

\frac{f(1) − f(0)}{1 − 0} = 1 − (−2) = 3

so we solve the equation 3x^{2}+2 = 3, or 3x^{2} = 1 or x^{2} = 1/3, i.e. x = ±1/\sqrt{3}. We care only for values c in the interval (0, 1), so we have only c = 1/\sqrt{3}.

Problem 2. Let f(x) = x^{2}/(x − 2). Show that there is no value of c such that

f(3) − f(1) = f′(c)(3 − 1).

Does this contradict the Mean Value Theorem? Why or why not?

Solution. We have f(3) − f(1) = 9 − (−1) = 10, so we want to show there is no c such that f′(c) = 5. Well,

f′(x) = \frac{(x − 2)(2x) − x^{2}}{(x − 2)^{2}} = \frac{x^{2} − 4x}{x^{2} − 4x + 4} = 5

which becomes

x^{2} − 4x = 5(x^{2} − 4x + 4) = 5x^{2} − 20x + 20
0 = 4x^{2} − 16x + 20 = 4(x^{2} − 4x + 5).

Applying the quadratic formula, we get

x = \frac{4 ± \sqrt{16 − 20}}{2}

therefore the original quadratic does not have any real roots. Therefore no such c exists.

The does not contradict the Mean Value Theorem because the original function is discontinuous at x = 2.

QUIZ #8: CALCULUS 1A (Stankova) Wednesday, March 17, 2004 Section 11:00–12:00 (Voight)

Problem 1. Let f(x) = xe^{x}.

(a) On what intervals is f increasing or decreasing? (Open or closed intervals are acceptable.) Explain your work.

(b) Find the local maximum and minimum values of f. Explain.

(c) Find the intervals of concavity and the inflection points of f.

(d) Draw the graph of f.

Solution. For (a), we compute that

f′(x) = e^{x} + xe^{x} = (1 + x)e^{x}.

Note that e^{x} > 0 for all x. Therefore f′(x) > 0 and f is increasing for 1+x > 0, i.e. x > −1, and similarly f is decreasing for x < −1. Therefore x is decreasing on the interval (−∞, −1] and increasing on the interval [−1, ∞).

For (b), we see that f′(x) = (1+x)e^{x} = 0 so 1+x = 0, therefore x = −1 is the only critical point. We compute

f′′(x) = (1 + x)e^{x} + e^{x} = (2 + x)e^{x}

and f′′(−1) = e^{−1} > 0 so x = −1 is a local minimum. There is no local maximum.

For (c), we see again since e^{x} > 0 that f′′(x) > 0 for 2 + x > 0, or x > −2. Therefore f is concave upward on (−2, ∞) and concave downward on (−∞, −2). So x = −2 is an inflection point.

Finally we have the graph for (d):