Question

find the x-coordinates of all local maxima using the second derivative test.
g(x)=x^{9}-3x^{7}+2
if there are multiple values, give them separated by commas.

• do not write x =.
• if there are no local maxima, enter ∅.
• use the exact answer, no decimals.

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:

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