use the second derivative test to find the x - coordinates of all local minima given the following…

use the second derivative test to find the x - coordinates of all local minima given the following function.\n\np(x)=x^{9}+3x^{8}+2\n\nif there are multiple values, give them separated by commas. if there are no local minima, enter ∅.

use the second derivative test to find the x - coordinates of all local minima given the following function.\n\np(x)=x^{9}+3x^{8}+2\n\nif there are multiple values, give them separated by commas. if there are no local minima, enter ∅.

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime=nx^{n - 1}), we have (p^\prime(x)=9x^{8}+24x^{7}=3x^{7}(3x + 8)) Set (p^\prime(x)=0), then (3x^{7}(3x + 8)=0) Solving (3x^{7}(3x + 8)=0) gives (x = 0) or (x=-\frac{8}{3})

Step2: Find the second derivative

Differentiate (p^\prime(x)=9x^{8}+24x^{7}) using the power rule. (p^{\prime\prime}(x)=72x^{7}+168x^{6}=12x^{6}(6x + 14))

Step3: Evaluate the second derivative at critical points

  • For (x = 0): (p^{\prime\prime}(0)=12\times0^{6}(6\times0 + 14)=0). The second - derivative test is inconclusive at (x = 0).
  • For (x=-\frac{8}{3}): (p^{\prime\prime}\left(-\frac{8}{3}\right)=12\times\left(-\frac{8}{3}\right)^{6}\left(6\times\left(-\frac{8}{3}\right)+14\right)) (\left(-\frac{8}{3}\right)^{6}>0), and (6\times\left(-\frac{8}{3}\right)+14=-16 + 14=-2<0), so (p^{\prime\prime}\left(-\frac{8}{3}\right)<0) (this is a local maximum)

Answer:

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