identify the inflection points and local maxima and minima of the function below and its associated graph…

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.\n\n( y = \frac { 3 } { 11 } ( x ^ { 2 } - 9 ) ^ { \frac { 2 } { 3 } } )\n\nseparate answers as needed)\n\nb. there are no local minima.\n\nfind the open interval(s) on which the function is differentiable and is concave up. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the curve is concave up on the open interval(s)\n\n(simplify your answer. type your answer in interval notation. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\n\nb. the curve is never concave up

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.\n\n( y = \frac { 3 } { 11 } ( x ^ { 2 } - 9 ) ^ { \frac { 2 } { 3 } } )\n\nseparate answers as needed)\n\nb. there are no local minima.\n\nfind the open interval(s) on which the function is differentiable and is concave up. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the curve is concave up on the open interval(s)\n\n(simplify your answer. type your answer in interval notation. type an exact answer, using radicals as needed. use a comma to separate answers as needed.)\n\nb. the curve is never concave up

Answer

Explanation:

Step1: Find the first - derivative

Use the chain rule (y = \frac{3}{11}(x^{2}-9)^{\frac{2}{3}}). Let (u=x^{2}-9), then (y=\frac{3}{11}u^{\frac{2}{3}}). The derivative of (y) with respect to (u) is (y_{u}=\frac{3}{11}\times\frac{2}{3}u^{-\frac{1}{3}}=\frac{2}{11}u^{-\frac{1}{3}}), and the derivative of (u) with respect to (x) is (u_{x} = 2x). By the chain rule (y_{x}=\frac{4x}{11(x^{2}-9)^{\frac{1}{3}}}).

Step2: Find the second - derivative

Use the quotient rule (y=\frac{4x}{11(x^{2}-9)^{\frac{1}{3}}}), where (u = 4x), (u'=4) and (v = 11(x^{2}-9)^{\frac{1}{3}}), (v'=\frac{11\times2x}{3(x^{2}-9)^{\frac{2}{3}}}). By the quotient rule (y''=\frac{4\times11(x^{2}-9)^{\frac{1}{3}}-4x\times\frac{11\times2x}{3(x^{2}-9)^{\frac{2}{3}}}}{121(x^{2}-9)^{\frac{2}{3}}}). Simplify the numerator: [ \begin{align*} &44(x^{2}-9)^{\frac{1}{3}}-\frac{88x^{2}}{3(x^{2}-9)^{\frac{2}{3}}}\ =&\frac{132(x^{2}-9)-88x^{2}}{3(x^{2}-9)^{\frac{2}{3}}}\ =&\frac{132x^{2}-1188 - 88x^{2}}{3(x^{2}-9)^{\frac{2}{3}}}\ =&\frac{44x^{2}-1188}{3(x^{2}-9)^{\frac{2}{3}}}\ =&\frac{44(x^{2}-27)}{3(x^{2}-9)^{\frac{2}{3}}} \end{align*} ]

Step3: Find where (y''>0)

Set (y''>0), (\frac{44(x^{2}-27)}{3(x^{2}-9)^{\frac{2}{3}}}>0). Since ((x^{2}-9)^{\frac{2}{3}}>0) for (x\neq\pm3), we solve (x^{2}-27>0). (x^{2}-27>0) gives (x< - 3\sqrt{3}) or (x>3\sqrt{3}).

Answer:

A. The curve is concave up on the open interval(s) ((-\infty,-3\sqrt{3}),(3\sqrt{3},\infty))