Find the critical points of $f(x)=x^{3}-15 x^{2}+48 x$ 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? \[ x= \]
Solution
Step 1 ：Find the derivative of the function \(f(x)=x^{3}-15 x^{2}+48 x\), which is \(f'(x) = 3x^{2}-30x+48\)
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: \(f''(8) = 6*8-30 = 18 > 0\), 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: \(f''(2) = 6*2-30 = -18 < 0\), so \(x=2\) is a local maximum
Step 9 ：Find the local maximum of \(f\) by evaluating \(f(2) = 2^{3}-15*2^{2}+48*2 = -4\)
Step 10 ：\(\boxed{c=2,8}\)
Step 11 ：\(\boxed{\text{local maximum: } f(x)=-4}\)
Step 12 ：\(\boxed{x=2}\)
