Find interval(s) where the function f(x)=4x^3-4x^2 +1 is increasing.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The interval where the function `f(x) = 4x^3 - 4x^2 +1` is increasing in the interval where f'(x) > 0

`f'(x) = 12x^2 - 8x`

`12x^2 - 8x > 0`

=> `4x(3x - 2) > 0`

This is true when `4x > 0` and `3x - 2 > 0`

=> `x > 0` and `x > 2/3`

=> `x > 2/3`

And when `4x < 0` and `3x - 2 < 0`

=> `x < 0` and `x < 2/3`

=> `x < 0`

The interval where the function is increasing is `(-oo, 0)U(2/3, oo)`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the first derivative to tell where the function increases or decreases such that:

`f'(x) = 12x^2 - 8x`

You need to solve for x the equation `f'(x) = 0`  such that:

`12x^2 - 8x = 0 => 4x(3x - 2) = 0 => 4x = 0 => x = 0`

`3x - 2 = 0 => 3x = 2 => x = 2/3`

You should know that between the roots, the values of `f'(x)`  are negative and outside the roots, `f'(x)>0` .

Since the sign of derivative of the function  decides if the function decreases or increases, hence, if f'(x) is positive, then, the function increases and if `f'(x)`  is negative, the function decreases.

Hence, the function increases if `x in (-oo,0)U(2/3,oo)`  and it decreases if `x in (0,2/3).`

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salonigaba | Student, Grade 11 | (Level 1) Salutatorian

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 f(x)=4x^3-4x^2 +1

 

Differnetiation: f'(x)=12x^2-8x

for increament or dicreament f'(x)=0

f'(x)=12x^2-8x=0

4x(3x-2)=0

x=0,2/3

x is positive it means it will increase.

Sources:
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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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I am sorry I some how had seen 4 x^3 - 4 x^2 + 1 as 4 x^3 , 4 x^2 + 1. But my procedure is correct so basically you just need to check f'(x)>0 region.  

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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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function:

f(x) = 4 x^3 is increasing from -infinity up to +infinity. just plot and check that the curve is going down and down when you move along the negative direction of "x" at x = 0 it is 0 and it goes on increasing for x>0. So its always increasing.

More mathematical way of answering is:

check where the first derivative of the function is positive:

f(x) = 4 x^2 so

f'(x) =  12 x^2, as x is squared so its always positive from -infinity to + infinity so the function f(x) is increasing in the interval " -infinity to +infinity". 

 

f(x) = 4 x^2 + 1 is increasing from 0 up to infinity. If you plot it you will see that the curve is symmetric about y-axis (f(x) axis) so its dicreasing on the left of y (or x=0) and increasing on the right of y (x=0), that means the function f(x) = 4 x^2 + 1 is increasing from "0" up to +infinity. 

Mathematically:

f'(x) = 8 x, which is -ve for x <0 and positive for x>0, so the function f(x) = 4 x^2 + 1  is increasing in the interval 0 to +infinity.

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