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 = 9) with output value 438. this function has a local maximum at (x = 2) with output value 105. question help: video 1 video 2 video 3 message instructor

Explanation:

Step1: Find the derivative of the function

Given $f(x)=2x^{3}-33x^{2}+108x + 5$, using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=6x^{2}-66x + 108$.

Step2: Set the derivative equal to zero

Set $f^\prime(x)=0$, so $6x^{2}-66x + 108 = 0$. Divide through by 6: $x^{2}-11x + 18=0$.

Step3: Factor the quadratic equation

Factor $x^{2}-11x + 18$ as $(x - 2)(x - 9)=0$.

Step4: Solve for x

Setting each factor equal to zero gives $x=2$ or $x = 9$.

Step5: Use the second - derivative test

Find the second - derivative $f^{\prime\prime}(x)=12x-66$.
When $x = 2$, $f^{\prime\prime}(2)=12\times2-66=24 - 66=-42<0$, so $f(x)$ has a local maximum at $x = 2$.
When $x = 9$, $f^{\prime\prime}(9)=12\times9-66=108 - 66 = 42>0$, so $f(x)$ has a local minimum at $x = 9$.

Step6: Find the local maximum and minimum values

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

Answer:

The function $f(x)=2x^{3}-33x^{2}+108x + 5$ has a local maximum at $x = 2$ with output value $105$ and a local minimum at $x = 9$ with output value $- 238$.