Find the critical points of $f(x)=x^3-15x^2+48x$ and use the Second Derivative Test (if possible) to determine whether each corresponds to a local minimum or maximum.
(Use symbolic notation and fractions where needed. Give your answer in the form of a comma separated list. Enter DNE if there are no critical points.)
$$c=$$

Find the local maximum of $f$.
$$\text{ local maximum: }f(x)=$$

At which point does the local maximum occur?
$$x=$$

Step 1 ：Find the derivative of the function $f(x)=x^3-15x^2+48x$, which is $f^{\prime }(x)=3x^2-30x+48$

Step 2 ：Set the derivative equal to zero: $3x^2-30x+48=0$

Step 3 ：Simplify the equation by dividing through by 3: $x^2-10x+16=0$

Step 4 ：Factor the equation: $(x-8)(x-2)=0$

Step 5 ：Set each factor equal to zero to find the critical points: $x=8,2$

Step 6 ：Find the second derivative of the function $f(x)$, which is $f^{″}(x)=6x-30$

Step 7 ：Evaluate $f^{″}(x)$ at $x=8$ to determine if it is a local minimum or maximum: $f^{″}(8)=6\ast 8-30=18>0$, so $x=8$ is a local minimum

Step 8 ：Evaluate $f^{″}(x)$ at $x=2$ to determine if it is a local minimum or maximum: $f^{″}(2)=6\ast 2-30=-18<0$, so $x=2$ is a local maximum

Step 9 ：Find the local maximum of $f$ by evaluating $f(2)=2^3-15\ast 2^2+48\ast 2=-4$

Step 10 ：$\boxed{{\displaystyle c=2,8}}$

Step 11 ：$\boxed{{\displaystyle \text{local maximum: }f(x)=-4}}$

Step 12 ：$\boxed{{\displaystyle x=2}}$