# Show function f(x)=(2x-m)/(2x^2+2x-1-m^2) don't have extrema if m real?

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### 1 Answer

You should prove that the given function has no extrema if `m in R` , hence, you need to prove that the equation `f'(x) = 0` has no solutions, under the given conditions, such that:

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

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

`f'(x) = (4x^2 + 4x - 2 - 2m^2 - 8x^2 - 4x + 4mx + 2m)/((2x^2+2x-1-m^2)^2)`

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

You need to prove that `f'(x)!=0` if `m in R` , hence, you need to prove that `b^2 - 4ac < 0` , thus, you need to identify the coefficients a,b,c, such that:

`a = -4, b = 4m, c = - 2m^2 + 2m - 2`

`b^2 - 4ac = 16m^2 - 16(2m^2 - 2m + 2) < 0`

`-16m^2 + 32m - 32 < 0 => m^2 - 2m + 1 > 0 => (m - 1)^2 > 0`

You should notice that the assumption that `b^2 - 4ac = 16m^2 - 16(2m^2 - 2m + 2) < 0` is valid, hence, the equation `f'(x) = 0` has no solutions for m in R.

**Hence, since the equation `f'(x) = 0` has no solutions for m in R, the function has no extrema, under the given conditions.**