dentify the inflection points and local maxima and\ninima of the function below and its associated\ngraph…

dentify the inflection points and local maxima and\ninima of the function below and its associated\ngraph. identify the open intervals on which the\nunction is differentiable and is concave up and\nconcave down.\n\n( y = \frac { 3 } { 7 } ( x ^ { 2 } - 25 ) ^ { \frac { 2 } { 3 } } )\n\nfind the inflection points of the curve. select the correct choice below and, if necessary, fill in the answer box\nto complete your choice.\na. the inflection point(s) is/are\n(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate\nanswers as needed.)\nb. there are no inflection points.

dentify the inflection points and local maxima and\ninima of the function below and its associated\ngraph. identify the open intervals on which the\nunction is differentiable and is concave up and\nconcave down.\n\n( y = \frac { 3 } { 7 } ( x ^ { 2 } - 25 ) ^ { \frac { 2 } { 3 } } )\n\nfind the inflection points of the curve. select the correct choice below and, if necessary, fill in the answer box\nto complete your choice.\na. the inflection point(s) is/are\n(type an ordered pair. type an exact answer, using radicals as needed. use a comma to separate\nanswers as needed.)\nb. there are no inflection points.

Answer

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}}}).

Step2: Find the second derivative

Use the quotient rule. If (y'=\frac{4x}{7(x^{2}-25)^{\frac{1}{3}}}), where (u = 4x), (u'=4) and (v = 7(x^{2}-25)^{\frac{1}{3}}), (v'=\frac{7\times2x}{3(x^{2}-25)^{\frac{2}{3}}}). By the quotient rule (y''=\frac{4\times7(x^{2}-25)^{\frac{1}{3}}-4x\times\frac{14x}{3(x^{2}-25)^{\frac{2}{3}}}}{49(x^{2}-25)^{\frac{2}{3}}}). Simplify the numerator: [ \begin{align*} &4\times7(x^{2}-25)^{\frac{1}{3}}-4x\times\frac{14x}{3(x^{2}-25)^{\frac{2}{3}}}\ =&\frac{84(x^{2}-25)-56x^{2}}{3(x^{2}-25)^{\frac{2}{3}}}\ =&\frac{84x^{2}-2100 - 56x^{2}}{3(x^{2}-25)^{\frac{2}{3}}}\ =&\frac{28x^{2}-2100}{3(x^{2}-25)^{\frac{2}{3}}}\ =&\frac{28(x^{2}-75)}{3(x^{2}-25)^{\frac{2}{3}}} \end{align*} ] Set (y'' = 0), then (x^{2}-75=0), so (x=\pm5\sqrt{3}).

Step3: Check the concavity

  • When (x\in(-\infty,- 5\sqrt{3})), let (x=-6\sqrt{3}), (y''=\frac{28((-6\sqrt{3})^{2}-75)}{3((-6\sqrt{3})^{2}-25)^{\frac{2}{3}}}>0)
  • When (x\in(-5\sqrt{3},5\sqrt{3})), let (x = 0), (y''=\frac{28(0 - 75)}{3(0 - 25)^{\frac{2}{3}}}<0)
  • When (x\in(5\sqrt{3},\infty)), let (x = 6\sqrt{3}), (y''=\frac{28((6\sqrt{3})^{2}-75)}{3((6\sqrt{3})^{2}-25)^{\frac{2}{3}}}>0)

Answer:

A. The inflection point(s) is/are ((-5\sqrt{3},\frac{3}{7}(75 - 25)^{\frac{2}{3}}), (5\sqrt{3},\frac{3}{7}(75 - 25)^{\frac{2}{3}}))