Consider the following function.
\[
f(x)=2 x^{3}-12 x^{2}+18 x-6
\]
Find the derivative.
\[
f^{\prime}(x)=6 x^{2}-24 x+18
\]
Find any critical numbers of the function. (Enter your answer as a comma-separated list. If any answer does not exist, enter DNE)
\[
x=1,3
\]
Find the interval(s) on which $f$ is increasing. (Enter your answer using interval notation.)
Find the interval(s) on which $f$ is decreasing. (Enter your answer using interval notation.)
Find the local minimum and maximum value of $f$.
local minimum value
local maximum value

Step 1 ：The derivative of the function \(f(x)=2 x^{3}-12 x^{2}+18 x-6\) is \(f^{\prime}(x)=6 x^{2}-24 x+18\).

Step 2 ：To find the critical numbers of the function, we set the derivative equal to zero and solve for \(x\). So, \(6 x^{2}-24 x+18=0\). Factoring out a 6, we get \(x^{2}-4 x+3=0\). Factoring further, we get \((x-1)(x-3)=0\). Setting each factor equal to zero gives us the critical numbers \(x=1,3\).

Step 3 ：To find the intervals on which \(f\) is increasing, we test the sign of the derivative on the intervals \((-\infty,1)\), \((1,3)\), and \((3,\infty)\). We find that \(f^{\prime}(x)>0\) on \((1,3)\) and \(f^{\prime}(x)<0\) on \((-\infty,1)\) and \((3,\infty)\). Therefore, \(f\) is increasing on \((1,3)\) and decreasing on \((-\infty,1)\) and \((3,\infty)\).

Step 4 ：To find the local minimum and maximum values of \(f\), we evaluate \(f\) at the critical numbers and at the endpoints of the domain. We find that \(f(1)=2\), \(f(3)=-6\), \(f(-\infty)=-\infty\), and \(f(\infty)=\infty\). Therefore, the local minimum value of \(f\) is \(-6\) and the local maximum value of \(f\) is \(2\).