# For `f(x)=1/(4x^4)-1/(3x^3)+2x` , use analytic methods-exact intervals: function increasing, decreasing, local extreme values. how do you factor y' for this question?

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You should find first derivative of the function using quotient rule, such that:

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

`f'(x) = -16x^3/(16x^8) + 9x^2/(9x^6) + 2`

Reducing duplicate factors yields:

`f'(x) = -1/(x^5) + 1/(x^4) + 2 => f'(x) = (-1 + x + 2x^5)/x^5`

You need to find the roots of derivative such that:

`f'(x) = 0 => (-1 + x + 2x^5)/x^5 = 0 => -1 + x + 2x^5 = 0`

You need to move the constant term to the right side such that:

`x + 2x^5 = 1`

You need to solve the equation using graphical method, hence, notice that the graph of the equation `y = x+2x^5` and `y = 1` meet each other at `x in (0,1).`

Hence, the function has an extreme point at `x_0 in (0,1).`

**You should notice that if `x = 0` , the derivative `f'(x)` is negative and if `x = 1` , the derivative is positive, hence the function decreases over `(-oo,x_0)` and it increases over `(x_0,oo), ` hence, the function reaches its local minimum at `x = x_0 in (0,1).` **