## Calculus with Applications (10th Edition)

$$f(x)=x^{2}-8 \ln x ; \quad\quad[1,4]$$ The critical number in the given interval is: $2$, The absolute maximum, $4.9096$ occurs when $x=4$ and the absolute minimum, $-1.54517$ occurs when $x=2.$
$$f(x)=x^{2}-8 \ln x ; \quad\quad[1,4]$$ First look for critical numbers in the interval $(1,4)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=(2)x-8(\frac{1}{x})\\ &=2x-\frac{8}{x}\\ &=\frac{2x^{2}-8}{x}\\ \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =\frac{2x^{2}-8}{x}=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & 2x^{2}-8=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & 2(x^{2}-4)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (x-2)(x+2)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & (x-2)=0 \quad \text {or} \quad(x+2)=0 \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=2 \quad \text {or} \quad x=-2 \\ \end{aligned} We know that $x=2 \in [1,4]$ but $-2 \notin [1,4]$ So, critical number in the given interval is: $2$, also $f^{\prime}(x)$ is not exist when $x=0$ but $0 \notin [1,4]$, so $x=0$ is not critical numbers in the given interval. Now, evaluate the function at the critical numbers and the endpoints: \begin{aligned} \text{when } x=1: \\ f(1) &=(1)^{2}-8 \ln (1)\\ &=(1)^{2}-8 (0) \\ &=1 \end{aligned} \begin{aligned} \text{when } x=2: \\ f(2) &=(2)^{2}-8 \ln (2)\\ &=4-8\ln \left(2\right)\\ &\approx -1.54517 \end{aligned} and \begin{aligned} \text{when } x=4: \\ f(4) &=(4)^{2}-8 \ln (4)\\ &=16-16\ln \left(2\right)\\ &\approx 4.9096 \end{aligned} We conclude that, the absolute maximum, $4.9096$ occurs when $x=4$ and the absolute minimum, $-1.54517$ occurs when $x=2.$