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 inflection points
find the local maxima. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the local maximum/maxima 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 maxima

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 \(y_{x}=\frac{dy}{du}\cdot\frac{du}{dx}\), so \(y'=\frac{2}{7}(x^{2}-25)^{-\frac{1}{3}}\cdot2x=\frac{4x}{7(x^{2}-25)^{\frac{1}{3}}}\).
Set \(y' = 0\), then \(4x=0\) gives \(x = 0\).
When \(x = 0\), \(y=\frac{3}{7}(0 - 25)^{\frac{2}{3}}=\frac{3}{7}\times25^{\frac{2}{3}}=\frac{3}{7}\times(5^{2})^{\frac{2}{3}}=\frac{3}{7}\times5^{\frac{4}{3}}=\frac{3\times5\times5^{\frac{1}{3}}}{7}=\frac{15\sqrt[3]{5}}{7}\).

Step2: Analyze the sign of the first derivative

For \(y'=\frac{4x}{7(x^{2}-25)^{\frac{1}{3}}}\), the critical points are \(x=- 5,x = 0,x = 5\).

• When \(x\lt - 5\), \(y'\lt0\) (e.g., \(x=-6\), \(y'=\frac{4\times(-6)}{7((-6)^{2}-25)^{\frac{1}{3}}}=\frac{-24}{7(11)^{\frac{1}{3}}}\lt0\)).
• When \(-5\lt x\lt0\), \(y'\gt0\) (e.g., \(x = - 1\), \(y'=\frac{4\times(-1)}{7((-1)^{2}-25)^{\frac{1}{3}}}=\frac{-4}{7(-24)^{\frac{1}{3}}}\gt0\)).
• When \(0\lt x\lt5\), \(y'\lt0\) (e.g., \(x = 1\), \(y'=\frac{4\times1}{7((1)^{2}-25)^{\frac{1}{3}}}=\frac{4}{7(-24)^{\frac{1}{3}}}\lt0\)).
• When \(x\gt5\), \(y'\gt0\) (e.g., \(x = 6\), \(y'=\frac{4\times6}{7((6)^{2}-25)^{\frac{1}{3}}}=\frac{24}{7(11)^{\frac{1}{3}}}\gt0\)).

Since the function changes from increasing (\(-5\lt x\lt0\)) to decreasing (\(0\lt x\lt5\)), there is a local maximum at \(x = 0\).

Answer:

A. The local maximum/maxima occur(s) at \((0,\frac{15\sqrt[3]{5}}{7})\)