MATH 180 FALL 1995

ASSIGNMENT #30

5.2, #17. The behavior of f(x) = e^{1/x} depends on the behavior of 1/x at the given limits. We find

\begin{matrix}x → ∞ \\ x → −∞ \\ x → 0^{+} \\ x → 0^{−} \end{matrix} \begin{matrix}1/x → 0 \\ 1/x → 0 \\ 1/x → ∞ \\ 1/x → −∞ \end{matrix} \begin{matrix}f(x) → 1 \\ f(x) → 1 \\ f(x) → ∞ \\ f(x) → 0 \end{matrix}

f′(x) = −\frac{1}{x^{2}} e^{1/x}, f′′(x) = \frac{1 + 2x}{x^{4}}e^{1/x}. Since f′(x) is always negative, f is decreasing on its domain. f′′(x) is zero at x = −1/2, f′′(x) is negative if x < −1/2, and f′′(x) is positive if x > −1/2 and x ≠ 0. (The point x = 0 is not in the domain of f). So f is concave down on (−∞, −1/2) and concave-up on (−1/2, 0) and (0, ∞).

5.2, #23. The graph should that of a decreasing function which alternates between concaveup and concave down in the four intervals determined by the points x_{1}, x_{2}, x_{3}.

5.3, #2. p(x) = x^{3} − ax, p′(x) = 3x^{2} − a, p′′(x) = 6x. If a < 0, then p′(x) = 3x^{2} − a is always positive and p(x) is always increasing. If a > 0, then p′(x) = 3x^{2} − a has roots at ±\sqrt{a/3}. Since p′ is positive for x < −\sqrt{a/3}, negative for −\sqrt{a/3} < x < \sqrt{a/3}, and positive for x > \sqrt{a/3}, it follows p has a local maximum at x = −\sqrt{a/3}, a local minimum at x = \sqrt{a/3}. p′′(x) changes sign at x = 0. So p is concave-down for x < 0 and concave-up for x > 0. The graphs are the cubics (a) and (c) illustrated in figure 5.34 on page 265 of the text.

5.3, #3. The discussion in the preceding problem shows p has a local maximum at x = −\sqrt{a/3} and a local minimum at x = \sqrt{a/3}. Now

f(−\sqrt{a/3}) = (2a/3)\sqrt{a/3}), f(\sqrt{a/3}) = −(2a/3)\sqrt{a/3}).

As a increases, the critical point (−\sqrt{a/3}, (2a/3)\sqrt{a/3}) moves left and up, the critical point (\sqrt{a/3}, −(2a/3)\sqrt{a/3}) moves right and down.

5.3, #4. f(x) = x^{2} + 2ax, f′(x) = 2x + 2a. The one critical point (−a, −a^{2}) of f is a local minimum. Indeed, (−a, −a^{2}) is the vertex of the parabola f(x). The roots of f(x) are x = −2a and x = 0 (this is easily ssen by factoring f(x) = x(x + 2a)). As a increases, the vertex (−a, −a^{2}) moves from right to left along the graph of y = −x^{2}; the root x = −2a moves from right to left; the root x = 0 stays fixed.