find a) any critical values and b) any relative extrema. g(x)= -x³ + 12x - 18 a) select the correct choice…

Explanation:

Step1: Find the derivative

The derivative of $g(x)= -x^{3}+12x - 18$ 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, since $x^{2}-4=(x + 2)(x - 2)$, we have $-3(x + 2)(x - 2)=0$. Solving for $x$, we get $x=-2$ or $x = 2$. 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 $x=-2$. $g(-2)=-(-2)^{3}+12\times(-2)-18=8-24 - 18=-34$. When $x = 2$, $g''(2)=-6\times2=-12<0$. So, the function has a relative maximum at $x = 2$. $g(2)=-(2)^{3}+12\times2-18=-8 + 24-18=-2$.

Answer:

a) A. The critical value(s) of the function is/are $-2,2$. b) The relative minimum is $(-2,-34)$ and the relative maximum is $(2,-2)$.