# y = f(x) f(x) + 2f(-x) = (x+1) : (x+2) What is f '(1) = ?

Asked on by blltrkmn

thilina-g | College Teacher | (Level 1) Educator

Posted on

I am assuming that you have indicated,

`f(x)+2f(-x) = (x+1)/(x+2)`

If `f(x)+2f(-x) = (x+1)/(x+2)` then,

`f(-x)+2f(-(-x)) = (-x+1)/(-x+2)`

This gives,

`f(-x)+2f(x) = (x-1)/(x-2)`

This gives,

`f(-x) = (x-1)/(x-2) - 2f(x)`

Therefore,

`f(x) + 2((x-1)/(x-2) - 2f(x)) = (x+1)/(x+2)`

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

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

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

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

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

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

Differentiating wrt x.

`3f'(x) = ((x^2-4)xx(2x+3) - (x^2+3x-2)xx2x)/(x^2-4)^2`

`3f'(x) = (2x^3+3x^2-8x-12 - (2x^3+6x^2-4x))/(x^2-4)^2`

`3f'(x) = (-3x^2-4x-12)/(x^2-4)^2`

`3f'(x) =- (3x^2+4x+12)/(x^2-4)^2`

`3f'(1) = -(3+4+12)/(1-4)^2`

`3f'(1) = -19/9`

`f'(1) = -19/27`

Therefore f'(1) = -19/27

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