let f(x)=(x - 2)^(2/3). find all points of absolute minima and maxima on the interval -3,4. as your answer…

Explanation:

Step1: Find the derivative

First, use the chain - rule. If $y=(x - 2)^{\frac{2}{3}}$, let $u=x - 2$, then $y = u^{\frac{2}{3}}$. The derivative of $y$ with respect to $u$ is $y^\prime_u=\frac{2}{3}u^{-\frac{1}{3}}$, and the derivative of $u$ with respect to $x$ is $u^\prime_x = 1$. So, $f^\prime(x)=\frac{2}{3}(x - 2)^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x - 2}}$. The derivative is undefined at $x = 2$.

Step2: Evaluate the function at critical and end - points

Evaluate $f(x)$ at the critical point $x = 2$ and the endpoints $x=-3$ and $x = 4$. When $x=-3$, $f(-3)=(-3 - 2)^{\frac{2}{3}}=5^{\frac{2}{3}}=\sqrt[3]{25}\approx2.924$. When $x = 2$, $f(2)=(2 - 2)^{\frac{2}{3}}=0$. When $x = 4$, $f(4)=(4 - 2)^{\frac{2}{3}}=2^{\frac{2}{3}}=\sqrt[3]{4}\approx1.587$.

Step3: Identify minima and maxima

The absolute minimum value of $f(x)$ on the interval $[-3,4]$ is $0$ (at $x = 2$), and the absolute maximum value is $\sqrt[3]{25}\approx2.924$ (at $x=-3$).

Step4: Calculate the sum

The sum of the values of $f$ at the points of absolute minima and maxima is $0 + 2.924=2.924$.

Answer:

$2.924$