find a) any critical values and b) any relative extrema. g(x)=x³ - 27x + 57. the critical value(s) of the…

find a) any critical values and b) any relative extrema. g(x)=x³ - 27x + 57. the critical value(s) of the function is/are. (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 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.) b. the relative minimum point(s) is/are (3,∞) and the relative maximum point(s) is/are (-∞,-3) (simplify your answers. type ordered pairs, using integers or fractions. use a comma to separate answers as needed.) c. 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.) d. there are no relative minimum points and there are no relative maximum points.
Answer
Explanation:
Step1: Find the derivative
The derivative of $g(x)=x^{3}-27x + 57$ using the power - rule $(x^n)'=nx^{n - 1}$ is $g'(x)=3x^{2}-27$.
Step2: Find critical values
Set $g'(x) = 0$. So, $3x^{2}-27=0$. First, factor out 3: $3(x^{2}-9)=0$, then $x^{2}-9=(x + 3)(x - 3)=0$. Solving $(x + 3)(x - 3)=0$ gives $x=-3,3$.
Step3: Use the second - derivative test
Find the second - derivative $g''(x)=6x$. When $x=-3$, $g''(-3)=6\times(-3)=-18<0$. So, the function has a relative maximum at $x = - 3$. $g(-3)=(-3)^{3}-27\times(-3)+57=-27 + 81+57=111$. The relative maximum point is $(-3,111)$. When $x = 3$, $g''(3)=6\times3=18>0$. So, the function has a relative minimum at $x = 3$. $g(3)=3^{3}-27\times3+57=27-81 + 57=3$. The relative minimum point is $(3,3)$.
Answer:
a) $-3,3$ b) The relative minimum point(s) is/are $(3,3)$ and the relative maximum point(s) is/are $(-3,111)$