# `f(x) = 4x^3 + 3x^2 - 6x + 1` (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...

`f(x) = 4x^3 + 3x^2 - 6x + 1` (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 verify where f'(x)>0 or f'(x)<0.

Hence, you need to differentiate the function with respect to x, such that:

`f'(x) = (4x^3 + 3x^2 - 6x + 1)'`

`f'(x) = 12x^2 + 6x - 6`

You need to set the equation f'(x) = 0, such that:

`12x^2 + 6x - 6 = 0`

Dividing by 6 yields:

`2x^2 + x - 1 = 0`

Re-write the equation `2x^2 + x - 1 = 0` such as:

`x^2 + x^2 + x - 1 = 0`

Group the terms, such that:

`(x^2 - 1) + (x^2 + x) = 0`

You need to write the difference of squares `x^2 - 1` such that:

`(x - 1)(x + 1) + (x^2 + x) = 0`

You need to factor out x in the group `x^2 + x` , such that:

`(x - 1)(x + 1) + x(x + 1) = 0`

Factor out x + 1:

`(x + 1)(x - 1 + x) = 0`

Put `x + 1 = 0 => x = -1`

Put `2x - 1 = 0 => 2x = 1 => x = 1/2`

You need to notice that f'(x)<0 for `x in (-1,1/2)` and f'(x)>0 for ` `

`x in (-oo,-1)U(1/2,+oo). `

Hence, the function decreases for `x in (-1,1/2)` and it increases for `x in (-oo,-1)U(1/2,+oo).`

b) The local maximum and minimum of the function are reached at values of x for f'(x) = 0.

From previous point a) yields that f'(x) = 0 for `x = -1` and `x = 1/2,` hence, the function reaches it's maximum at `x = -1` and it's minimum at `x = 1/2.`

Hence, the maximum of the function is the point `(-1,f(-1)) ` and the minimum is the point `(1/2,f(1/2)).`

c) The function is concave up if f''(x)>0 and it is concave down if f''(x)<0. You need to evaluate the second derivative:

`f''(x) = (12x^2 + 6x - 6)'`

`f''(x) = 24x + 6`

You need to put f''(x) = 0:

`24x + 6 = 0`

Dividing by 6 yields:

`4x + 1 = 0`

`x = -1/4`

Hence, the function has an inflection point at `x = -1/4` and it is concave down for `x in (-oo,1/4)` and it is concave up if `x in (1/4,oo).```

scisser | Student

find the derivative of the function

`f'(x)=12x^2+6x-6 `

Set it equal to 0

`12x^2+6x-6=0`

`x=-1/2 and x=-1`

Plugging in values to determine increasing or decreasing, we get `f'(-2)=30` , which would be increasing, and `f'(0)=-6` , which would be decreasing.

For inflection points, find the second derivative and set it equal to 0

`24x+6=0`

`x=-1/4`

Check for values around the point for inflection. `f''(0)=6` and `f''(-1) = -18` , which are concave up and concave down, respectively.

The function is increasing over `(-oo,-1)` and `(1/2,oo)` and decreasing over `(-1,1/2)` .

There is a local max at `x=-1` and local min at `x=1/2` . Lastly, there is an inflection point at `x=-1/4`