let f(x)=(x - 2)^(2/5). find all points of absolute minima and maxima on the interval -9,4. as your answer…

let f(x)=(x - 2)^(2/5). find all points of absolute minima and maxima on the interval -9,4. as your answer, input the sum of the values of f at all points the absolute minima and maxima occur in the given interval. your answer needs to be entered in decimal form with three significant digits after the decimal point.

let f(x)=(x - 2)^(2/5). find all points of absolute minima and maxima on the interval -9,4. as your answer, input the sum of the values of f at all points the absolute minima and maxima occur in the given interval. your answer needs to be entered in decimal form with three significant digits after the decimal point.

Answer

Explanation:

Step1: Find the derivative of $f(x)$

Using the chain - rule, if $y = u^{\frac{2}{5}}$ and $u=x - 2$, then $\frac{dy}{du}=\frac{2}{5}u^{-\frac{3}{5}}$ and $\frac{du}{dx}=1$. So $f^\prime(x)=\frac{2}{5}(x - 2)^{-\frac{3}{5}}=\frac{2}{5(x - 2)^{\frac{3}{5}}}$.

Step2: Find the critical points

The derivative $f^\prime(x)$ is undefined when $x = 2$ (since the denominator is zero at $x = 2$). There are no values of $x$ for which $f^\prime(x)=0$ because the numerator of $f^\prime(x)$ is a non - zero constant ($2$).

Step3: Evaluate the function at the critical point and endpoints

Evaluate $f(x)$ at $x=-9$, $x = 2$, and $x = 4$. When $x=-9$, $f(-9)=(-9 - 2)^{\frac{2}{5}}=(-11)^{\frac{2}{5}}=\sqrt[5]{121}\approx2.62$. When $x = 2$, $f(2)=(2 - 2)^{\frac{2}{5}}=0$. When $x = 4$, $f(4)=(4 - 2)^{\frac{2}{5}}=2^{\frac{2}{5}}=\sqrt[5]{4}\approx1.32$.

Step4: Determine the absolute minima and maxima

The absolute minimum value of $f(x)$ on the interval $[-9,4]$ is $0$ (at $x = 2$), and the absolute maximum value is $\sqrt[5]{121}\approx2.62$ (at $x=-9$).

Answer:

$2.620$