Question

use the second derivative test for local extrema to find the x - values of all local extrema for the given function.
f(x)=x^{3}-12x + 4
does the function have any local minimums? select the correct choice, and, if necessary, fill in the answer box(es) to complete your choice.
a. the function f(x) has one local minimum, located at x=
(simplify your answer.)
b. the function f(x) has two local minimums, located at x=
and x=
(simplify your answers.)
c. the function f(x) has three local minimums, located at x=
,x=
, and x=
(simplify your answers.)
d. the function f(x) has no local minimums.

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 \(f^\prime(x)=0\), so \(3x^{2}-12 = 0\). Factor out 3: \(3(x^{2}-4)=0\), then \(x^{2}-4=(x + 2)(x - 2)=0\). Solving gives \(x=-2\) 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

Evaluate \(f^{\prime\prime}(x)\) at critical points:

• When \(x=-2\), \(f^{\prime\prime}(-2)=6\times(-2)=-12<0\). So \(x = - 2\) is a local maximum.
• When \(x = 2\), \(f^{\prime\prime}(2)=6\times2=12>0\). So \(x = 2\) is a local minimum.

Answer:

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