## Calculus with Applications (10th Edition)

$$f(x)=\frac{1-x}{3+x}; \quad\quad[0,3]$$ The graph has no critical numbers. The absolute maximum, $\frac{1}{3} \approx 0.3333$ occurs when $x=0,$ and the absolute minimum, $\frac{-1}{3} \approx -0.3333$ occurs when $x=3.$
$$f(x)=\frac{1-x}{3+x}; \quad\quad[0,3]$$ First look for critical numbers in the interval $(0,3)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=\frac{1-x}{3+x};\\ &=\frac{(3+x) (-1)-(1-x)(1)}{(3+x)^{2}}\\ &=\frac{-4}{(3+x)^{2}} \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x) &=\frac{-4}{(3+x)^{2}}=0 \end{aligned} Since this equation has no solution, So,the graph has no critical numbers. Now, evaluate the function at the endpoints: $$\text{when } x=0: f(0)=\frac{1-(0)}{3+(0)}=\frac{1}{3} \approx 0.3333\\ \text{when } x=3: f(3)=\frac{1-(3)}{3+(3)}=\frac{-2}{6}\approx -0.3333\\$$ We conclude that, the absolute maximum, $\frac{1}{3} \approx 0.3333$ occurs when $x=0$ and the absolute minimum, $\frac{-1}{3} \approx -0.3333$ occurs when $x=3.$