## Calculus with Applications (10th Edition)

$$f(x)=\frac{\ln x}{ x^{2}} ; \quad\quad[1,4]$$ The critical number in the given interval is: $e ^{\frac{1}{2}}$. The absolute maximum, $0.18393$ occurs when $x=e ^{\frac{1}{2}}$ and the absolute minimum, $0$ occurs when $x=1.$
$$f(x)=\frac{\ln x}{ x^{2}} ; \quad\quad[1,4]$$ First look for critical numbers in the interval $(1,4)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=\frac{x^{2}.(\frac{1}{x}) -\ln x . (2x)}{ x^{4}} \\ &=\frac{x - 2x\ln x }{ x^{4}} \\ &=\frac{x(1- 2\ln x) }{ x^{4}}\\ &=\frac{(1- 2\ln x) }{ x^{3}}\\ \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =\frac{(1- 2\ln x) }{ x^{3}}=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (1- 2\ln x)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & 2\ln x=1\\ \Rightarrow\quad\quad\quad\quad\quad\\ & \ln x=\frac{1}{2}\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=e ^{\frac{1}{2}} \approx 1.64872\\ \end{aligned} So, the critical number in the given interval is: $e ^{\frac{1}{2}}$, also $f^{\prime}(x)$ is not exist when $x=0$ but $0 \notin [1,4]$, so $x=0$ is not critical number in the given interval. Now, evaluate the function at the critical numbers and the endpoints: \begin{aligned} \text{when } x=1: \\ f(1) &=\frac{\ln (1)}{ (1)^{2}} \\ &=\frac{0}{ 1} \\ &=0 \end{aligned} \begin{aligned} \text{when } x=e ^{\frac{1}{2}}: \\ f(e ^{\frac{1}{2}}) &=\frac{\ln (e ^{\frac{1}{2}})}{ (e ^{\frac{1}{2}})^{2}} \\ &=\frac{1}{2e}\\ &\approx 0.18393 \end{aligned} and \begin{aligned} \text{when } x=4: \\ f(4) &=\frac{\ln (4)}{ (4)^{2}} \\ &=\frac{\ln \left(2\right)}{8} \\ &\approx 0.08664 \end{aligned} We conclude that, the absolute maximum, $0.18393$ occurs when $x=e ^{\frac{1}{2}}$ and the absolute minimum, $0$ occurs when $x=1.$