# f(x)=3x^4-4x^3. where f is increasing? decreasing?concave up? concave down? infliction points? Given `f(x)=3x^4-4x^3` :

Note that a function is increasing if the first derivative is positive and decreasing where the first derivative is negative. All extrema occur only at critical points (the first derivative is zero or fails to exist.)

The function is concave up when the second derivative is positive,...

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Given `f(x)=3x^4-4x^3` :

Note that a function is increasing if the first derivative is positive and decreasing where the first derivative is negative. All extrema occur only at critical points (the first derivative is zero or fails to exist.)

The function is concave up when the second derivative is positive, concave down when the second derivative is negative. Inflection points occur where the second derivative is zero and changes sign.

(1) `f'(x)=12x^3-12x^2` Setting the first derivative equal to zero we get:

`12x^3-12x^2=0`

`12x^2(x-1)=0`

So the first derivative is zero when x=0 and x=1.

On `(-oo,0)` `f'(x)<0` (Try a test point like x=-1; `f'(-1)=-24` )

` ` On `(0,1)`   `f'(x)<0` (Try a test point like x=1/2; `f'(1/2)=-1.5` )

On (1,0) `f'(x)>0`

So the function is decreasing on `(-oo,1)` and increasing on `(1,oo)`

(2) `f''(x)=36x^2-24x`

`36x^2-24x=0`

`12x(3x-2)=0`

`x=0`  or `x=2/3`

On `(-oo,0)` `f''(x)>0`

On `(0,2/3)` `f''(x)<0`

On `(2/3,oo)` `f''(x)>0`

So the function is concave up on `(-oo,0)` and `(2/3,oo)` , concave down on `(0,2/3)` with an inflection points at x=0 and `x=2/3`

The graph:

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