Calculus problem, look below

The function f(x) = 2 x^3 + 9 x^2 - 240 x - 1 is decreasing on the interval ( , ).
It is increasing on the interval ( negavtive infty, ) and the interval ( , infty ).
The function has a local maximum at . Fill in the blanks

Lol is this an online homework problem? I'm only doing this because I'm also a calc student and I need the extra practice :P
Anyway:
Find the derivative so you get
6x^2 + 18x - 240 = 0
6 (x^2 + 3x - 40) = 0
6 (x+8) (x-5) = 0
Critical Points: -8, 5

Increasing (neg. Inf, -8) and (5, inf)
Decreasing (-8, 5)

firs t you take the derivative

and then you take the second derivative


f ' = 6x^2 + 18 x - 240

f '' = 12x + 18


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f ' = 0 at local max and mins

0 = 6x^2 + 18 x - 240

0 = x^2 + 3x - 40

find the factors .. what adds to 3 and multiples to - 40

8 and -5 ... are the factors

0 = (x + 8) (x - 5)

x + 8 = 0 or x -5 = 0

x = -8

x = 5

so the max and mins are at - 8 or 5

use f '' to find if its max or min

f '' < 0 for a max

f '' > 0 for a min

f '' = 12x + 18

f '' = 12*5 + 18 > 0

f '' = 12*(-8) + 18 < 0

last question is local max at x = -8

then use f(x) = 2 x^3 + 9 x^2 - 240 x - 1 to find the y value by putting x = -8 into the equation

I will leave that for you its a simple math problem..

the point will be for the local max ( -8, f( -8))

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we can find when f (x) is decreasing by solving the f ' equation

f (x) is decreasing when the slope is negative...

f ' < 0

f ' = 6x^2 + 18 x - 240 < 0

6x^2 + 18x - 240 < 0

x^2 + 3x - 40 < 0

(x +8)(x - 5) < 0