# Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 47

(a) $f(x)$ has critical point at $x = e^{−2}$. (b) There is a local minimum at $x = \frac{1}{e^2}$. (c) $x = \frac{1}{e^2}$ yields an absolute minimum of $f(1/e^2) = −\frac{2}{e}≈ −.736$. There is no absolute maximum since $f$ increases without bound as $x→∞$.

#### Work Step by Step

(a). $\frac{\sqrt{x}}{x}+\frac{ln(x)}{2\sqrt{x}}+\frac{1+2 n(x)}{2\sqrt{x}}$ . This is defined everywhere on $(0,∞)$ and is $0$ only at $x = e^{−2}$. (b). Note that $f < 0$ on $(0, \frac{1}{e^2})$ and $f > 0$ on $(\frac{1}{e^2},∞)$, so there is a local minimum at $x = \frac{1}{e^2}$. (c). Since there is only one critical point, the local minimum at $x = \frac{1}{e^2}$ yields an absolute minimum of $f(1/e^2) = −\frac{2}{e}≈ −.736$. There is no absolute maximum since $f$ increases without bound as $x→∞$.

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