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

Explanation:

Step1: Find the derivative

$f'(x)=6x^{2}-66x + 108$

Step2: Set the derivative equal to 0

$6x^{2}-66x + 108 = 0$
Divide through by 6: $x^{2}-11x + 18=0$
Factor: $(x - 2)(x - 9)=0$
Solve for $x$: $x = 2$ or $x=9$

Step3: Use the second - derivative test

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

Step4: Find the function values

$f(2)=2\times2^{3}-33\times2^{2}+108\times2 + 5=2\times8-33\times4 + 216+5=16-132 + 216+5=105$
$f(9)=2\times9^{3}-33\times9^{2}+108\times9 + 5=2\times729-33\times81+972 + 5=1458-2673+972 + 5=-238$

Answer:

This function has a local minimum at $x = 9$ with output value: $-238$
and a local maximum at $x = 2$ with output value: $105$