use the second derivative test for local extrema to find the x - values of all local extrema for the given…

use the second derivative test for local extrema to find the x - values of all local extrema for the given function.\nf(x)=x^{3}-12x + 4\n(simplify your answers.)\nd. the function f(x) has no local minimums.\ndoes the function have any local maximums? select the correct choice, and, if necessary, fill in the answer box(es) to complete your choice.\na. the function f(x) has one local maximum, located at x=\n(simplify your answer.)\nb. the function f(x) has two local maximums, located at x= and x=\n(simplify your answers.)\nc. the function f(x) has three local maximums, located at x=, x=, and x=\n(simplify your answers.)\nd. the function f(x) has no local maximums.

use the second derivative test for local extrema to find the x - values of all local extrema for the given function.\nf(x)=x^{3}-12x + 4\n(simplify your answers.)\nd. the function f(x) has no local minimums.\ndoes the function have any local maximums? select the correct choice, and, if necessary, fill in the answer box(es) to complete your choice.\na. the function f(x) has one local maximum, located at x=\n(simplify your answer.)\nb. the function f(x) has two local maximums, located at x= and x=\n(simplify your answers.)\nc. the function f(x) has three local maximums, located at x=, x=, and x=\n(simplify your answers.)\nd. the function f(x) has no local maximums.

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (y = f(x)=x^{3}-12x + 4), the first derivative (f^\prime(x)=3x^{2}-12). Set (f^\prime(x)=0), then (3x^{2}-12 = 0). Factor out 3: (3(x^{2}-4)=0), which is (3(x - 2)(x + 2)=0). Solving for (x), we get (x=-2) and (x = 2).

Step2: Find the second derivative

Differentiate (f^\prime(x)=3x^{2}-12) with respect to (x). Using the power rule, (f^{\prime\prime}(x)=6x).

Step3: Apply the second - derivative test

  • For (x=-2): Substitute (x=-2) into (f^{\prime\prime}(x)), (f^{\prime\prime}(-2)=6\times(-2)=-12<0). When (f^{\prime}(c) = 0) and (f^{\prime\prime}(c)<0), the function has a local maximum at (x = c). So (f(x)) has a local maximum at (x=-2).
  • For (x = 2): Substitute (x = 2) into (f^{\prime\prime}(x)), (f^{\prime\prime}(2)=6\times2 = 12>0). When (f^{\prime}(c)=0) and (f^{\prime\prime}(c)>0), the function has a local minimum at (x = c).

Answer:

A. The function (f(x)) has one local maximum, located at (x=-2)