# Determine when f(x) `<=` g(x) if f(x) = `(x^2+3x-4)/(2x-1)` andg(x)=` -(x-4)/(x+2) `

*print*Print*list*Cite

First we need to find the points of intersection.

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

Cross multiplying we get

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

` x^3+3x^2-4x+2x^2+6x-8=-(2x^2-8x-x+4)`

`x^3+5x^2+2x-8=-2x^2+9x-4`

Putting into standard form we get

`x^3+7x^2-7x-4=0`

Roots are x=1.2406793, x=-0.411814527, x=-7.82886477

x>1.2406793 f(2) =(2^2+3(2)-4)/(2(2)-1)=(4+6-4)/(4-1)=6/3 = 2, g(2)=-(2-4)/(2+2) = -(-2)/4 = 1/2 so when x>1.2406793 f(x)>=g(x)

1.2406793 >= x >= 1/2 f(1) = (1^2+3(1)-4)/(2(1)-1)=(1+3-4)/(2) = 0, g(1)=-(1-4)/(1+2) = -(-3)/3 = 1 so f(x)<=g(x)

We also need to look before and after any points where f(x) or g(x) is undefined f(x) is undefined at -1/2 and g(x) is undefined at (-2).

1/2 > x >= -0.411814527 f(0) = (-4)/(-1) = 4, g(0) = -(-4)/(2) = 2, so f(x) >= g(x).

-2 > x >= -0.411814527 f(-1) = ((-1)^2+3(-1)-4)/(2(-1)-1)=(1-3-4)/(-2-1)=(-6)/(-3)=2, g(-1) = -(-1-4)/(-1+2) = -(-5)/1 = 5 so f(x) <= g(x)

f(-3) = ((-3)^2+3(-3)-4)/(2(-3)-1)=(9-9-4)/(-6-1)=(-4)/(-7) = 4/7, g(-3) = -(-3-4)/(-3+2) = -(-7)/(-1)=-7 so f(x)>=g(x)

f(-10) = ((-10)^2+3(-10)-4)/(2(-10)-1) = (100-30-4)/(-21) = -66/21, g(-10)=-(-10-4)/(-10+2) = -(-14)/(-8) = -7/8 so f(x) <= g(x).

So in conclusion:

f(x)<=g(x) when -0.411814527 > x >-2, and 1/2 < x >= 1.2406793