Answer
(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 $x→∞$.
