Question

consider the function ( q(x)=x^{11}-3 x^{10}+2 ). use the second derivative test to find the ( x )-coordinates of all local maxima. if there are multiple values, give them separated by commas. if there are no local maxima, enter ( varnothing ). provide your answer below:

Explanation:

Step1: Find the first derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(q(x)=x^{11}-3x^{10}+2\), the first derivative \(q^\prime(x)=11x^{10}-30x^{9}=x^{9}(11x - 30)\)

Set \(q^\prime(x) = 0\), then \(x^{9}(11x - 30)=0\). Solving \(x^{9}=0\) gives \(x = 0\), and solving \(11x-30=0\) gives \(x=\frac{30}{11}\approx2.73\)

Step2: Find the second derivative

Differentiate \(q^\prime(x)=11x^{10}-30x^{9}\) again. Using the power rule, \(q^{\prime\prime}(x)=110x^{9}-270x^{8}=10x^{8}(11x - 27)\)

Step3: Evaluate the second derivative at critical points

• For \(x = 0\):

\(q^{\prime\prime}(0)=10\times0^{8}(11\times0 - 27)=0\). The second - derivative test is inconclusive for \(x = 0\)

• For \(x=\frac{30}{11}\):

\(q^{\prime\prime}(\frac{30}{11})=10\times(\frac{30}{11})^{8}(11\times\frac{30}{11}-27)=10\times(\frac{30}{11})^{8}(30 - 27)=30\times(\frac{30}{11})^{8}>0\)

Since \(q^{\prime\prime}(\frac{30}{11})>0\), \(x = \frac{30}{11}\) is a local minimum.

Answer:

\(\varnothing\)