find the x-coordinates of all local maxima using the second derivative test.\ng(x)=x^{9}-3x^{7}+2\nif there…

find the x-coordinates of all local maxima using the second derivative test.\ng(x)=x^{9}-3x^{7}+2\nif there are multiple values, give them separated by commas.\n• do not write x =.\n• if there are no local maxima, enter ∅.\n• use the exact answer, no decimals.

find the x-coordinates of all local maxima using the second derivative test.\ng(x)=x^{9}-3x^{7}+2\nif there are multiple values, give them separated by commas.\n• do not write x =.\n• if there are no local maxima, enter ∅.\n• use the exact answer, no decimals.

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), we have (g^\prime(x)=9x^{8}-21x^{6}=3x^{6}(3x^{2}-7)). Set (g^\prime(x) = 0), then (3x^{6}(3x^{2}-7)=0). From (x^{6}=0), we get (x = 0). From (3x^{2}-7=0), we get (x=\pm\sqrt{\frac{7}{3}}).

Step2: Find the second derivative

Differentiate (g^\prime(x)=9x^{8}-21x^{6}) again. Using the power rule, (g^{\prime\prime}(x)=72x^{7}-126x^{5}=18x^{5}(4x^{2}-7)).

Step3: Apply the second - derivative test

  • For (x = 0): (g^{\prime\prime}(0)=18\times0^{5}(4\times0^{2}-7)=0). The second - derivative test is inconclusive for (x = 0).
  • For (x=\sqrt{\frac{7}{3}}): (g^{\prime\prime}(\sqrt{\frac{7}{3}})=18(\sqrt{\frac{7}{3}})^{5}(4\times\frac{7}{3}-7)=18(\sqrt{\frac{7}{3}})^{5}\times\frac{7}{3}>0). So (x=\sqrt{\frac{7}{3}}) is a local minimum.
  • For (x =-\sqrt{\frac{7}{3}}): (g^{\prime\prime}(-\sqrt{\frac{7}{3}})=18(-\sqrt{\frac{7}{3}})^{5}(4\times\frac{7}{3}-7)=18(-\sqrt{\frac{7}{3}})^{5}\times\frac{7}{3}<0).

Answer:

(-\sqrt{\frac{7}{3}})