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Consider the following function.
$$f(x)=x^3-3x^2-6x+3$$
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 $f$.
$$(x,y)=(◻)$$
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Step 1 ：The first derivative of the function is found by applying the power rule, which gives $f^{\prime }(x)=3x^2-6x-6$.

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 $c=\boxed{{\displaystyle 1}}$.

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 $(1,\mathrm{\infty })$.

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 $(1,-5)$.