Find a) any critical values and b) any relative extrema.
$$g(x)=x^3-27x+53$$
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 $-3,3$.
(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 $◻$ 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.)
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.

Step 1 ：First, we need to find the derivative of the function $g(x)=x^3-27x+53$. The derivative of a function at a certain point gives the slope of the tangent line at that point. Critical points occur where the derivative is zero or undefined.

Step 2 ：The derivative of $g(x)$ is $g^{\prime }(x)=3x^2-27$.

Step 3 ：To find the critical points, we set the derivative equal to zero and solve for $x$. So, $3x^2-27=0$.

Step 4 ：Dividing through by 3 gives $x^2-9=0$.

Step 5 ：Factoring the equation gives $(x-3)(x+3)=0$.

Step 6 ：Setting each factor equal to zero gives $x=3$ and $x=-3$. So, the critical values of the function are $x=3$ and $x=-3$. Therefore, the answer to part a) is A. The critical value(s) of the function is/are -3,3.

Step 7 ：Next, we need to determine whether these critical points are relative extrema. We do this by using the second derivative test. The second derivative of a function at a certain point gives the curvature of the function at that point. If the second derivative is positive, the function is concave up and the point is a relative minimum. If the second derivative is negative, the function is concave down and the point is a relative maximum.

Step 8 ：The second derivative of $g(x)$ is $g^{″}(x)=6x$.

Step 9 ：Substituting $x=3$ into the second derivative gives $g^{″}(3)=18$, which is positive. Therefore, $x=3$ is a relative minimum.

Step 10 ：Substituting $x=-3$ into the second derivative gives $g^{″}(-3)=-18$, which is negative. Therefore, $x=-3$ is a relative maximum.

Step 11 ：To find the y-coordinates of these points, we substitute $x=3$ and $x=-3$ into the original function $g(x)$.

Step 12 ：Substituting $x=3$ into $g(x)$ gives $g(3)=3^3-27\ast 3+53=-2$. So, the relative minimum point is $(3,-2)$.

Step 13 ：Substituting $x=-3$ into $g(x)$ gives $g(-3)=(-3{)}^3-27\ast (-3)+53=98$. So, the relative maximum point is $(-3,98)$.

Step 14 ：Therefore, the answer to part b) is B. The relative minimum point(s) is/are $(3,-2)$ and the relative maximum point(s) is/are $(-3,98)$.