Question

the function $f(x)=2x^{3}-33x^{2}+108x + 5$ has one local minimum and one local maximum. use a graph of the function to estimate these local extrema. this function has a local minimum at $x =$ with output value: and a local maximum at $x =$ with output value: question help: video message instructor post to forum submit question

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=2x^{3}-33x^{2}+108x + 5$ is $f'(x)=6x^{2}-66x + 108$.

Step2: Set the derivative equal to zero

$6x^{2}-66x + 108 = 0$. Divide through by 6: $x^{2}-11x + 18=0$.

Step3: Factor the quadratic equation

$(x - 2)(x - 9)=0$. So the critical - points are $x = 2$ and $x=9$.

Step4: Use the second - derivative test

The second - derivative $f''(x)=12x-66$.
For $x = 2$, $f''(2)=12\times2-66=-42<0$, so $x = 2$ is a local maximum.
For $x = 9$, $f''(9)=12\times9-66=42>0$, so $x = 9$ is a local minimum.

Step5: Find the output values

$f(2)=2\times2^{3}-33\times2^{2}+108\times2 + 5=2\times8-33\times4 + 216+5=16-132 + 216+5=105$.
$f(9)=2\times9^{3}-33\times9^{2}+108\times9 + 5=2\times729-33\times81+972 + 5=1458-2673+972 + 5=-438$.

Answer:

This function has a local minimum at $x = 9$ with output value: $-438$
and a local maximum at $x = 2$ with output value: $105$