Given the function $f(x)=x^{3}-3 x^{2}-24 x+11$ over the interval $-6 \leq x \leq 5$, answer the following questions.
(a) Find $f^{\prime}(x)$ and $f^{\prime \prime}(x)$.
\[
\begin{array}{l}
f^{\prime}(x)= \\
f^{\prime \prime}(x)=
\end{array}
\]
(b) Find the critical points of $f$.
\[
x_{1}=
\]
\[
x_{2}=
\]
(c) Find any inflection points of $f$.
\[
x=
\]
(d) Evaluate $f$ at its critical points and at the endpoints of the given interval. Identify local and global maxima and minima of $f$ in the interval.
\[
\begin{array}{l}
f\left(x_{1}\right)= \\
f\left(x_{2}\right)= \\
f(-6)= \\
f(5)=
\end{array}
\]

Step 1 ：Take the first derivative of \(f(x)\) to find \(f^{\prime}(x)\): \(f^{\prime}(x)=3x^{2}-6x-24\)

Step 2 ：Take the second derivative of \(f(x)\) to find \(f^{\prime \prime}(x)\): \(f^{\prime \prime}(x)=6x-6\)

Step 3 ：Find the critical points of \(f\) by setting \(f^{\prime}(x)=0\): \(3x^{2}-6x-24=0\)

Step 4 ：Simplify the equation to find the critical points: \(x^{2}-2x-8=0\)

Step 5 ：Factor the equation to find the critical points: \((x-4)(x+2)=0\)

Step 6 ：Set each factor equal to zero to find the critical points: \(x_{1}=4\) and \(x_{2}=-2\)

Step 7 ：Find the inflection points of \(f\) by setting \(f^{\prime \prime}(x)=0\): \(6x-6=0\)

Step 8 ：Solve for \(x\) to find the inflection point: \(x=1\)

Step 9 ：Evaluate \(f\) at its critical points and at the endpoints of the given interval to identify local and global maxima and minima: \(f(4)=-33\), \(f(-2)=27\), \(f(-6)=-1\), and \(f(5)=-39\)

Step 10 ：The local maximum is at \(x=-2\) with a value of \(27\), the local minimum is at \(x=4\) with a value of \(-33\), the global maximum is at \(x=-6\) with a value of \(-1\), and the global minimum is at \(x=5\) with a value of \(-39\)