## Calculus with Applications (10th Edition)

$$f(x)=\frac{8+x}{8-x}; \quad\quad[4,6]$$ The graph has no critical numbers. The absolute maximum, $7$ occurs when $x=6$ and the absolute minimum, $3$ occurs when $x=4.$
$$f(x)=\frac{8+x}{8-x}; \quad\quad[4,6]$$ First look for critical numbers in the interval $(4,6)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=\frac{8+x}{8-x};\\ &=\frac{(8-x) (1)-(8+x)(-1)}{(8-x)^{2}}\\ &=\frac{16}{(8-x)^{2}} \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x) &=\frac{16}{(8-x)^{2}}=0 \end{aligned} Since this equation has no solution, also $f^{\prime}(x)$is not exist when $x=8$ but $8 \notin [4,6]$, so the graph has no critical numbers. Now, evaluate the function at the endpoints: $$\text{when } x=4: f(4)=\frac{8+(4)}{8-(4)} =3\\ \text{when } x=6: f(6)=\frac{8+(6)}{8-(6)} =7\\$$ We conclude that, the absolute maximum, $7$ occurs when $x=6$ and the absolute minimum, $3$ occurs when $x=4.$