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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)`
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).`
for increament or dicreament f'(x)=0
x is positive it means it will increase.
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.
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.
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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