12. find the absolute maximum and minimum of $f(x)=5x^{4/5}+4x$ on $-2,1$.

Explanation:

Step1: Find the derivative

Using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=5\times\frac{4}{5}x^{-\frac{1}{5}}+4 = 4x^{-\frac{1}{5}}+4=\frac{4}{x^{\frac{1}{5}}}+4$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. Then $\frac{4}{x^{\frac{1}{5}}}+4 = 0$. Subtract 4 from both sides: $\frac{4}{x^{\frac{1}{5}}}=-4$. Cross - multiply to get $4=-4x^{\frac{1}{5}}$. Then $x^{\frac{1}{5}}=-1$, and $x=-1$. Also, $f^\prime(x)$ is undefined at $x = 0$. So the critical points are $x=-1$ and $x = 0$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at $x=-2,x=-1,x = 0,x = 1$.

• When $x=-2$, $f(-2)=5(-2)^{\frac{4}{5}}+4\times(-2)=5\times2^{\frac{4}{5}}-8$.
• When $x=-1$, $f(-1)=5(-1)^{\frac{4}{5}}+4\times(-1)=5 - 4=1$.
• When $x = 0$, $f(0)=5\times0^{4/5}+4\times0 = 0$.
• When $x = 1$, $f(1)=5\times1^{4/5}+4\times1=5 + 4=9$.

Answer:

The absolute maximum value of $f(x)$ on $[-2,1]$ is 9 (at $x = 1$) and the absolute minimum value is $5\times2^{\frac{4}{5}}-8$ (at $x=-2$).