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}{11}left(x^{2}-9
ight)^{\frac{2}{3}} )

separate 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. Let \(u = x^{2}-9\), then \(y=\frac{3}{11}u^{\frac{2}{3}}\).
The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=\frac{3}{11}\times\frac{2}{3}u^{-\frac{1}{3}}=\frac{2}{11}(x^{2} - 9)^{-\frac{1}{3}}\), and the derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx}=2x\).
By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{4x}{11(x^{2}-9)^{\frac{1}{3}}}\).
Set \(\frac{dy}{dx} = 0\), then \(4x=0\), so \(x = 0\).

Step2: Find the second derivative

Use the quotient rule \((\frac{f}{g})^\prime=\frac{f^\prime g - fg^\prime}{g^{2}}\), where \(f = 4x\) and \(g=11(x^{2}-9)^{\frac{1}{3}}\).
\(f^\prime=4\) and \(g^\prime=\frac{11\times2x}{3(x^{2}-9)^{\frac{2}{3}}}\).
\(\frac{d^{2}y}{dx^{2}}=\frac{4\times11(x^{2}-9)^{\frac{1}{3}}-4x\times\frac{22x}{3(x^{2}-9)^{\frac{2}{3}}}}{121(x^{2}-9)^{\frac{2}{3}}}=\frac{132(x^{2}-9)-88x^{2}}{363(x^{2}-9)^{\frac{4}{3}}}=\frac{44x^{2}-1188}{363(x^{2}-9)^{\frac{4}{3}}}=\frac{4(x^{2}-27)}{33(x^{2}-9)^{\frac{4}{3}}}\).

Step3: Analyze the first - derivative for local maxima/minima

We use the first - derivative test.
When \(x\lt0\) (e.g., \(x=-1\)), \(\frac{dy}{dx}=\frac{- 4}{11(-1 + 9)^{\frac{1}{3}}}=\frac{-4}{11\times2}\lt0\).
When \(x\gt0\) (e.g., \(x = 1\)), \(\frac{dy}{dx}=\frac{4}{11(1 + 9)^{\frac{1}{3}}}\gt0\).
Since the function changes from decreasing (\(\frac{dy}{dx}\lt0\)) to increasing (\(\frac{dy}{dx}\gt0\)) at \(x = 0\).

Step4: Calculate the \(y\) - value at \(x = 0\)

Substitute \(x = 0\) into \(y=\frac{3}{11}(x^{2}-9)^{\frac{2}{3}}\), we get \(y=\frac{3}{11}(0 - 9)^{\frac{2}{3}}=\frac{3}{11}\times9^{\frac{2}{3}}=\frac{3}{11}\times(3^{2})^{\frac{2}{3}}=\frac{3}{11}\times3^{\frac{4}{3}}=\frac{3^{\frac{7}{3}}}{11}\).

Answer:

A. The local maximum/maxima occur(s) at \((0,\frac{3^{\frac{7}{3}}}{11})\)