Question

question 11
the function ( f(x)=2 x^{3}-39 x^{2}+216 x + 4 ) has one local minimum and one local maximum. use a graph of the function to estimate these local extrema. round to the nearest whole number.
this function has a local minimum at ( x=) with output value
and a local maximum at ( x=) with output value

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=2x^{3}-39x^{2}+216x + 4\) is \(f^{\prime}(x)=6x^{2}-78x + 216\).
Factor out \(6\): \(f^{\prime}(x)=6(x^{2}-13x + 36)\).
Factor the quadratic: \(f^{\prime}(x)=6(x - 4)(x - 9)\).

Step2: Find the critical points

Set \(f^{\prime}(x)=0\), then \(6(x - 4)(x - 9)=0\).
Solving \(x-4 = 0\) gives \(x = 4\), and solving \(x - 9=0\) gives \(x=9\).

Step3: Use the second - derivative test

The second - derivative \(f^{\prime\prime}(x)=12x-78\).
For \(x = 4\): \(f^{\prime\prime}(4)=12\times4-78=48 - 78=-30<0\). So \(x = 4\) is a local maximum.
For \(x = 9\): \(f^{\prime\prime}(9)=12\times9-78=108 - 78 = 30>0\). So \(x = 9\) is a local minimum.

Step4: Calculate the function values

For \(x = 4\): \(f(4)=2\times4^{3}-39\times4^{2}+216\times4 + 4=2\times64-39\times16 + 864+4=128-624+864 + 4=372\).
For \(x = 9\): \(f(9)=2\times9^{3}-39\times9^{2}+216\times9+4=2\times729-39\times81+1944 + 4=1458-3159+1944+4=247\).

Answer:

This function has a local minimum at \(x = 9\) with output value \(247\) and a local maximum at \(x = 4\) with output value \(372\).