Question

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 -2,2 (use a comma to separate answers as needed.)
b. the function has no critical values.
b) select the correct choice below and, if necessary, fill in the answer box(es) within your choice
a. the relative minimum point(s) is/are and the relative maximum point(s) is/are (simplify your answers. type ordered pairs, using integers or fractions. use a comma to separate answers as needed.)
b. the relative minimum point(s) is/are and there are no relative maximum points. (simplify your answer. type an ordered pair, using integers or fractions. use a comma to separate answers as needed.)
c. the relative maximum point(s) is/are and there are no relative minimum points. (simplify your answer. type an ordered pair, using integers or fractions. use a comma to separate answers as needed.)
d. there are no relative minimum points and there are no relative maximum points

Explanation:

Step1: Find the derivative of g(x)

$g(x)=-x^{3}+12x - 10$, so $g'(x)=-3x^{2}+12$

Step2: Set the derivative equal to zero to find critical values

$-3x^{2}+12 = 0$
$3x^{2}=12$
$x^{2}=4$
$x=\pm2$

Step3: Use the second - derivative test to find relative extrema

Find the second - derivative $g''(x)=-6x$
When $x = 2$, $g''(2)=-6\times2=-12<0$, so there is a relative maximum at $x = 2$.
$g(2)=-2^{3}+12\times2 - 10=-8 + 24-10 = 6$
When $x=-2$, $g''(-2)=-6\times(-2)=12>0$, so there is a relative minimum at $x=-2$.
$g(-2)=-(-2)^{3}+12\times(-2)-10=8-24 - 10=-26$

Answer:

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