Question

identify the inflection points and local maxima and minima of the function below and its associated graph. identify the open intervals on which the function is differentiable and is concave up and concave down.

$y = \frac { 3 } { 7 } ( x ^ { 2 } - 25 ) ^ { \frac { 2 } { 3 } }$

answers as needed.)
b. there are no local maxima.
find the local minima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the local minimum/minima occur(s) at
(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)
b. there are no local minima

Explanation:

Step1: Find the first derivative

Use the chain rule. If \(y = \frac{3}{7}(x^{2}-25)^{\frac{2}{3}}\), let \(u=x^{2}-25\), then \(y=\frac{3}{7}u^{\frac{2}{3}}\).
The derivative of \(y\) with respect to \(u\) is \(y_{u}'=\frac{3}{7}\times\frac{2}{3}u^{-\frac{1}{3}}=\frac{2}{7}u^{-\frac{1}{3}}\), and the derivative of \(u\) with respect to \(x\) is \(u_{x}' = 2x\).
By the chain rule \(y'=\frac{2}{7}(x^{2}-25)^{-\frac{1}{3}}\times2x=\frac{4x}{7(x^{2}-25)^{\frac{1}{3}}}\).
Set \(y' = 0\), then \(4x=0\) gives \(x = 0\). Also, \(y'\) is undefined when \(x^{2}-25=0\), i.e., \(x=\pm5\).

Step2: Analyze critical points for local minima

We use the first - derivative test.

• For \(x<-5\), say \(x=-6\), \(y'=\frac{4\times(-6)}{7((-6)^{2}-25)^{\frac{1}{3}}}=\frac{-24}{7(11)^{\frac{1}{3}}}<0\).
• For \(-5 < x<0\), say \(x=-1\), \(y'=\frac{4\times(-1)}{7((-1)^{2}-25)^{\frac{1}{3}}}=\frac{-4}{7(-24)^{\frac{1}{3}}}>0\).
• For \(0 < x<5\), say \(x = 1\), \(y'=\frac{4\times1}{7(1 - 25)^{\frac{1}{3}}}=\frac{4}{7(-24)^{\frac{1}{3}}}<0\).
• For \(x>5\), say \(x=6\), \(y'=\frac{4\times6}{7(6^{2}-25)^{\frac{1}{3}}}=\frac{24}{7(11)^{\frac{1}{3}}}>0\).

At \(x=-5\), the function changes from decreasing (\(x < - 5\)) to increasing (\(-55\)).

When \(x=-5\), \(y=\frac{3}{7}((-5)^{2}-25)^{\frac{2}{3}}=0\). When \(x = 5\), \(y=\frac{3}{7}(5^{2}-25)^{\frac{2}{3}}=0\).

Answer:

A. The local minimum/minima occur(s) at \((-5,0),(5,0)\)