## Calculus with Applications (10th Edition)

$$f(x)=5x^{\frac{2}{3}}+2x^{\frac{5}{3}}; \quad\quad[-2,1]$$ The critical numbers in the given interval are: -1 and 0. The absolute maximum, $7$ occurs when $x=1$ and the absolute minimum, $0$ occurs when $x=0.$
$$f(x)=5x^{\frac{2}{3}}+2x^{\frac{5}{3}}; \quad\quad[-2,1]$$ First look for critical numbers in the interval $(-2,1)$. find the points where the derivative is $0$ \begin{aligned} f^{\prime}(x) &=5(\frac{2}{3})x^{\frac{-1}{3}}+2(\frac{5}{3})x^{\frac{2}{3}}\\ &=(\frac{10}{3})x^{\frac{-1}{3}}+(\frac{10}{3})x^{\frac{2}{3}}\\ &=(\frac{10}{3x^{\frac{1}{3}}})+(\frac{10x^{\frac{2}{3}}}{3})\\ &=\frac{10+10x}{3x^{\frac{1}{3}}}\\ &=\frac{10(1+x)}{3x^{\frac{1}{3}}}\\ \end{aligned} Solve the equation $f^{\prime}(x)=0$ to get \begin{aligned} f^{\prime}(x)& =\frac{10(1+x)}{3x^{\frac{1}{3}}}=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & 10(1+x)=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=-1 \end{aligned} The derivative $f^{\prime}(x)$ is undefined when \begin{aligned} 3x^{\frac{1}{3}}&=0\\ \Rightarrow\quad\quad\quad\quad\quad\\ & x=0 \end{aligned} Since $f(0)$ is defined, then $x=0$ is also critical point Therefore, the critical numbers in the given interval are: -1 and 0. Now, evaluate the function at the critical numbers and the endpoints: \begin{aligned} \text{when } x=-2: &\\ f(-2) &=5(-2)^{\frac{2}{3}}+2(-2)^{\frac{5}{3}}\\ &=2^{\frac{2}{3}}\\ &\approx 1.5874 \end{aligned} \begin{aligned} \text{when } x=-1: &\\ f(-1) &=5(-1)^{\frac{2}{3}}+2(-1)^{\frac{5}{3}}\\ &=3 \end{aligned} \begin{aligned} \text{when } x=0: &\\ f(0) &=5(0)^{\frac{2}{3}}+2(0)^{\frac{5}{3}}\\ &=0\\ \end{aligned} and \begin{aligned} \text{when } x=1: &\\ f(1) &=5(1)^{\frac{2}{3}}+2(1)^{\frac{5}{3}}\\ &=7 \end{aligned} We conclude that, the absolute maximum, $7$ occurs when $x=1$ and the absolute minimum, $0$ occurs when $x=0.$