Question

the function $f(x)=2x^{3}-36x^{2}+162x + 6$ 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:

Explanation:

Step1: Find the derivative

$f'(x)=6x^{2}-72x + 162$

Step2: Set the derivative equal to zero

$6x^{2}-72x + 162=0$
Divide through by 6: $x^{2}-12x + 27=0$
Factor: $(x - 3)(x - 9)=0$
Solve for $x$: $x=3$ or $x = 9$

Step3: Use the second - derivative test

$f''(x)=12x-72$
When $x = 3$, $f''(3)=12\times3-72=-36<0$, so $f(x)$ has a local maximum at $x = 3$.
When $x = 9$, $f''(9)=12\times9-72 = 36>0$, so $f(x)$ has a local minimum at $x = 9$.

Step4: Find the function values

$f(3)=2\times3^{3}-36\times3^{2}+162\times3 + 6=2\times27-36\times9 + 486+6=54-324 + 486+6=222$
$f(9)=2\times9^{3}-36\times9^{2}+162\times9 + 6=2\times729-36\times81+1458+6=1458-2916+1458+6=6$

Answer:

This function has a local minimum at $x = 9$ with output value: $6$
and a local maximum at $x = 3$ with output value: $222$