Question

If `f(x) = (x^2-4)/ (x-1)`, what is `f(0), f(1/2), f(-2), f(x-2), f(r^2) and f(1/t)`?

Answer 1

`f(0)=4`
`f(1/2)=15/2`
`f(-2) = 0`
`f(x-2)=(x(x-4))/(x-3)`
`f(r^2)=(r^4-4)/(r^2-1)`
`f(1/t)=((1/t^2)-4)/((1/t)-1)`

Explanation:

To find the expressions that replaces the x in f(x), we have to replace the x in the equation with whatever is in the bracket, so if

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

`f(0) = (0^2-4)/(0-1)= (-4)/(-1)=4`

`f(1/2) = ((1/2)^2-4)/((1/2)-1)=((1/4)-4)/((1/2)-1)=15/2`

`f(-2) = ((-2)^2-4)/((-2)-1)=(0)/-3=0`

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

`f(r^2) = ((r^2)^2-4)/((r^2)-1)=(r^4-4)/(r^2-1)`

`f(1/t) = ((1/t)^2-4)/((1/t)-1)=((1/t^2)-4)/((1/t)-1)`