consider the function ( q(x)=x^{11}-3 x^{10}+2 ). use the second derivative test to find the ( x…

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:

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