Find the local maximum and local minimum of the function. f(x) = x^3 − 12x^2 − 27x + 7

dy/dx = 3x^2 - 24x - 27

= 0 for a max/min

divide by 3
x^2 - 8x - 9 = 0
it factors nicely ...
(x-9)(x+1) = 0
x = 9 or x = -1

f(9) = .....
f(-1) = ....

take over.

this confirms our work
http://www.wolframalpha.com/input/?i=plot+f%28x%29+%3D+x%5E3+%E2%88%92+12x%5E2+%E2%88%92+27x+%2B+7

Those are the answers I got that I thought were wrong, but the online assignment wanted the value of f(9)=-479 and f(-1)=21, not 9 and -1. Thank you!

exactly.

I left it up to you to finish the problem.
Oh well ....

f(9) = 9^3 - 12(9)^2 - 27(9) + 7
= -479

etc for f(-1) = 21

btw, I also did your other very long question.

Did you look at it ?

To find the local maximum and local minimum of a function, we need to find the critical points of the function. A critical point is a point on the function where its derivative is either zero or does not exist. We can find the critical points by following these steps:

Step 1: Find the derivative of the function.
f'(x) = 3x^2 - 24x - 27

Step 2: Set f'(x) equal to zero and solve for x to find potential critical points.
3x^2 - 24x - 27 = 0

Step 3: Solve the quadratic equation to find the values of x.
Using the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 3, b = -24, and c = -27.
x = (-(-24) ± sqrt((-24)^2 - 4*3*(-27))) / (2*3)
x = (24 ± sqrt(576 + 324)) / 6
x = (24 ± sqrt(900)) / 6
x = (24 ± 30) / 6
x = (24 + 30) / 6 or x = (24 - 30) / 6
x = 54 / 6 or x = -6 / 6
x = 9 or x = -1

Step 4: Determine if these critical points are local maximum or local minimum.
To determine this, we can use the second derivative test.
f''(x) = 6x - 24

For x = 9:
f''(9) = 6(9) - 24 = 54 - 24 = 30
Since f''(9) > 0, the function has a local minimum at x = 9.

For x = -1:
f''(-1) = 6(-1) - 24 = -6 - 24 = -30
Since f''(-1) < 0, the function has a local maximum at x = -1.

Therefore, the local maximum is at x = -1, and the local minimum is at x = 9.