Question

find a) any critical values and b) any relative extrema. g(x)= -x^3 + 12x - 10 a) select the correct choice below and, if necessary, fill in the answer box within your choice a. the critical value(s) of the function is/are (use a comma to separate answers as needed.) b. the function has no critical values

Explanation:

Step1: Find the derivative

The derivative of $g(x)=-x^{3}+12x - 10$ using the power - rule $(x^n)'=nx^{n - 1}$ is $g'(x)=-3x^{2}+12$.

Step2: Set the derivative equal to zero

Set $g'(x)=0$, so $-3x^{2}+12 = 0$. First, factor out - 3: $-3(x^{2}-4)=0$. Then, using the difference of squares $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 2$, we have $-3(x + 2)(x - 2)=0$. Solving for $x$, we get $x=-2$ and $x = 2$. These are the critical values.

Step3: Use the second - derivative test for relative extrema

Find the second - derivative $g''(x)$. Differentiating $g'(x)=-3x^{2}+12$ gives $g''(x)=-6x$.
When $x=-2$, $g''(-2)=-6\times(-2)=12>0$. So, the function has a relative minimum at $x=-2$. Substitute $x=-2$ into $g(x)$: $g(-2)=-(-2)^{3}+12\times(-2)-10=8-24 - 10=-26$.
When $x = 2$, $g''(2)=-6\times2=-12<0$. So, the function has a relative maximum at $x = 2$. Substitute $x = 2$ into $g(x)$: $g(2)=-(2)^{3}+12\times2-10=-8 + 24-10=6$.

Answer:

a) A. The critical value(s) of the function is/are $-2,2$
b) The relative minimum is $-26$ at $x=-2$ and the relative maximum is $6$ at $x = 2$.