## Calculus with Applications (10th Edition)

$$f(x)=x^{2} e^{-0.5x} ; \quad\quad[2,5]$$ The critical number in the given interval is: $4$. The absolute maximum, $2.16536$ occurs when $x=4$ and the absolute minimum, $1.47151$ occurs when $x=2.$
$$f(x)=x^{2} e^{-0.5x} ; \quad\quad[2,5]$$ First look for critical numbers in the interval $(2,5)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=x^{2}\left(-\frac{1}{2} e^{-0.5 x}\right)+2 x e^{-0.5 x} \\ &=e^{-0.5 x}\left(2 x-\frac{1}{2} x^{2}\right) \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =e^{-0.5 x}\left(2 x-\frac{1}{2} x^{2}\right)=0\\ &=\frac{x}{2}e^{-0.5 x}\left(4 - x\right)=0 \Rightarrow\quad\quad\quad\quad\quad\\ & x=0 \quad \text {or} \quad(4-x)=0 \\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=0 \quad \text {or} \quad x=4 \\ \end{aligned} We know that $x=4 \in [2,5]$ but $0 \notin [2,5]$ So, the critical number in the given interval is: $4$. Now, evaluate the function at the critical numbers and the endpoints: \begin{aligned} \text{when } x=2: \\ f(2) &=(2)^{2} e^{-0.5(2)} \\ &=\frac{4}{e} \\ &\approx 1.47151 \end{aligned} \begin{aligned} \text{when } x=4: \\ f(4) &=(4)^{2} e^{-0.5(4)} \\ &=\frac{16}{e^2} \\ &\approx 2.16536 \end{aligned} and \begin{aligned} \text{when } x=5: \\ f(5) &=(5)^{2} e^{-0.5(5)} \\ &=5^2\cdot \frac{1}{e^{2.5}}\\ &\approx 2.05212 \end{aligned} We conclude that, the absolute maximum, $2.16536$ occurs when $x=4$ and the absolute minimum, $1.47151$ occurs when $x=2.$