# `f(x) = 2x^3 + 3x^2 - 36x` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the...

`f(x) = 2x^3 + 3x^2 - 36x` (a) Find the intervals on which `f` is increasing or decreasing. (b) Find the local maximum and minimum values of `f`. (c) Find the intervals of concavity and the inflection points.

sciencesolve | Certified Educator

a) You need to determine the monotony of the function, hence, you need to remember that f(x) increases if f'(x)>0 and f(x) decreases if f'(x)<0.

You need to evaluate f'(x), such that:

`f'(x) = 6x^2+6x-36`

Putting f'(x) = 0, yields:

`6x^2+6x-36 = 0 => x^2 + x - 6 = 0`

You need to use quadratic formula to evaluate the zeroes of the equation:

`x_(1,2) = (-1+-sqrt(1+24))/2 => x_(1,2) = (-1+-5)/2`

`x_1 = 2 ; x_2 = -3`

You need to notice that the function increases for `x in (-oo,-3)U(2,oo)` and it decreases if `x in (-3,2)` .

b) The function hasĀ  maximum and minimum points at x values such as f'(x) = 0.

From point a) yields that f'(x) = 0 for x = -3 and x = 2.

Hence, the function has a maximum point at (-3,f(-3)) and it has a minimum point at (2,f(2)).

c) The function is concave up for f''(x)>0 and it is concave down for f''(x)<0.

`f''(x) = 12x + 6`

Putting f''(x) = 0, yields:

`12x + 6 = 0 => x = -6/12 => x = -1/2`

Hence, the function has an inflection point at `(-1/2,f(-1/2))` and the function is concave up for `x in (-1/2,oo)` and it is concave down for `x in (-oo,-1/2).`

scisser | Student

(a)
`f(x) = 2x^3 + 3x^2 - 36x `
`f'(x) = 6x^2 + 6x - 36 `
`x = -3, 2 `
`f'(-10) = 600 - 60 - 36 gt 0` , so f is increasing on `(-oo,-3)` .
`f'(0) = 0 + 0 - 36 lt 0` , so f is decreasing on `(-3,2)` .
`f'(10) = 600 + 60 - 36 gt 0` , so f is increasing on `(2,oo)` .

(b)
Extrema occur when `f'(x) = 0` and `f'(x)` changes sign at x.
As we've seen, such extrema occur at -3 and 2.
As f is increasing to `f(-3)` and decreasing afterward, a maximum occurs here.
As f is decreasing to `f(2)` and increasing afterward, a minimum occurs here.
`f(-3) = 81` .
`f(2) = -44` .
So a minimum occurs at `(2,-44)` and a maximum occurs at `(-3,81)` .

(c)
`f''(x) = 12x + 6 `
`0 = 12x + 6 `
`x = -1/2 `
`f''(-10) = -120 + 6 lt 0` , so f is concave down on `(-oo,-1/2)` .
`f''(0) = 0 + 6 gt 0` , so f is concave up on `(-1/2,oo)` .

(d) graph the function