Question

How do you find interval of increasing, decreasing, concave up and down for `f(x) = 2x^3-3x^2-36x-7`?

Answer 1

The intervals of increasing are `x in (-oo,-2)uu(3,+oo)` and the interval of decreasing is `x in (-2,3)`. Please see below for the concavities.

Explanation:

The function is

`f(x)=2x^3-3x^2-36x-7`

To fd the interval of increasing and decreasing, calculate the first derivative

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

To find the critical points, let `f'(x)=0`

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

`=>`, `x^2-x-6=0`

`=>`, `(x-3)(x+2)=0`

The critical points are

`{(x=3),(x=-2):}`

Build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `↗` `color(white)(aaaa)` `↘` `color(white)(aaaa)` `↗`

The intervals of increasing are `x in (-oo,-2)uu(3,+oo)` and the interval of decreasing is `x in (-2,3)`

Calculate the second derivative

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

The point of inflection is when `f''(x)=0`

`=>`, `12x-6=0`

`=>`, `x=1/2`

The intervals to consider are `(-oo,1/2)` and `(1/2,+oo)`

Build a variation chart

`color(white)(aaaa)` `" Interval "` `color(white)(aaaa)` `(-oo,1/2)` `color(white)(aaaa)` `(1/2,+oo)`

`color(white)(aaaa)` `" sign f''(x) "` `color(white)(aaaaaa)` `(-)` `color(white)(aaaaaaaaaa)` `(+)`

`color(white)(aaaa)` `" f(x) "` `color(white)(aaaaaaaaaaa)` `nn` `color(white)(aaaaaaaaaaaa)` `uu`

The function is concave down in the interval `(-oo,1/2)` and concave down in the interval `(1/2,+oo)`.

graph{2x^3-3x^2-36x-7 [-26.64, 46.44, 1.46, 38]}