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Consider the following function.
\[
f(x)=x^{3}-3 x^{2}-6 x+3
\]
Find the first and second derivatives.
\[
\begin{array}{l}
f^{\prime}(x)= \\
f^{\prime \prime}(x)=
\end{array}
\]
Find any values of $c$ such that $f^{\prime \prime}(c)=0$. (Enter your answer as a comma-separated list. If any answer does not exist, enter DNE)
\[
c=
\]
Find the interval(s) on which $f$ is concave up. (Enter your answer using interval notation.)
Find the interval(s) on which $f$ is concave down. (Enter your answer using interval notation.)
Find the inflection point of $f$.
\[
(x, y)=(\square)
\]
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Step 1 ：The first derivative of the function is found by applying the power rule, which gives \(f'(x) = 3x^2 - 6x - 6\).

Step 2 ：The second derivative of the function is the derivative of the first derivative, which gives \(f''(x) = 6x - 6\).

Step 3 ：The values of \(c\) such that \(f''(c) = 0\) are found by setting the second derivative equal to zero and solving for \(x\), which gives \(c = \boxed{1}\).

Step 4 ：The intervals on which the function is concave up are found using the second derivative test. If the second derivative is positive at a point, the function is concave up at that point. This gives the interval \((1, \infty)\).

Step 5 ：The inflection point of the function is a point where the function changes concavity. This occurs where the second derivative equals zero and changes sign, which gives the inflection point at \((1, -5)\).