Please Help I dont get this math problem? Thank you.

The function f(x)=2x^(3)-27x^(2)+108x+5 has one local minimum and one local maximum. Use a graph of the function to estimate these local extrema.

The local maximum and the local minimum are points on the graph which have a gradient of 0.

I trust you know what a positive cubic graph looks like.

If you graph that on WolframAlpha.com, the local maximum (FOR THE GRAPH) is the left bump and the local minimum (FOR THE GRAPH) is the right bump.

I don't know why your teacher wants you to look it up on a graph. But good luck if you have to draw it.

If not, you can simply use differentiation to find it.

f'(x) = 6x^2 - 54x + 108 = 0 for max/min
x^2 - 9x + 18 = 0
(x-6)(x-3) = 0

So at x = 6 and x = 3, are turning points.

Now f''(x) = 12x - 54 = 0 (the second derivative is used as a test (not in all cases though) to distinguish between the local maximum and local minimum, where if f''(x) < 0, that point is a local maximum and if f''(x) > 0, that point is a local minimum.

So f''(6) = 72 - 54 = 18
f''(3) = 36 - 54 = -18

f(6) = 432 - 972 + 648 + 5 = 113
f(3) = 54 - 243 + 324 + 5 = 140

So (6,113) is a local minimum, (3, 140) is a local maximum.
If you look on WolframAlpha, I should be right.

@Chris 54 divided by 6 is not 7, it's 9.

f(x)=2x^(3)-27x^(2)+108x+5
f'(x) = 6x^2 - 54x + 108
f(x) = 6(x^2 - 9x + 18)
f(x) = 6(x - 6)(x - 3)
6(x - 6)(x - 3) = 0
x - 6 = 0 or x - 3 = 0
x = 6 or x = 3
f(6) = 432 - 972 + 648 + 5 = 113
f(3) = 54 - 243 + 324 + 5 = 140

So the local maximum, which occurs at x = 3, is 140.
And the local minimum, which occurs at x = 6, is 113.

We can't do this in text, obviously, so I'd recommend https://www.desmos.com/calculator
Remember, minimum is the least y value, and the maximum is the greatest value of y.

EDIT: This is for graphing, remember.

You have to use a graphing calculator to get the graph of the function.