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 1 video 2 video 3 message instructor submit question

Explanation:

Step1: Find the derivative

Differentiate $f(x)=2x^{3}-33x^{2}+108x + 5$ using the power - rule. The derivative $f'(x)=6x^{2}-66x + 108$.

Step2: Set the derivative equal to zero

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

Step3: Solve the quadratic equation

Factor the quadratic equation: $x^{2}-11x + 18=(x - 2)(x - 9)=0$. The solutions are $x = 2$ and $x=9$.

Step4: Use the second - derivative test

Differentiate $f'(x)$ to get the second - derivative $f''(x)=12x-66$.
When $x = 2$, $f''(2)=12\times2-66=24 - 66=-42<0$. So $f(x)$ has a local maximum at $x = 2$.
When $x = 9$, $f''(9)=12\times9-66=108 - 66 = 42>0$. So $f(x)$ has a local minimum at $x = 9$.

Step5: Find the function values

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

Answer:

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