Question

F:[1/2,infinity)—>[3/4,infinity)

f(x)=x^2-x+1
find the inverse of f(x);please explain

Answer 1

`f^(-1)(x)=(1)/(2)+\sqrt{x-(3)/(4)}`

Explanation:

y=x^2-x+1

We want to solve for x.

I'm going to use completing the square.

Subtract 1 on both sides:

y-1=x^2-x

Add (-1/2)^2 on both sides:

y-1+(-1/2)^2=x^2-x+(-1/2)^2

This allows me to write the right hand side as a square.

y-1+1/4=(x-1/2)^2

y-3/4=(x-1/2)^2

Now remember we are solving for x so now we square root both sides:

`\pm √(y-3/4)=x-1/2`

The problem said the domain was 1/2 to infinity and the range was 3/4 to infinity.

This is only the right side of the parabola because of the domain restriction. We want x-1/2 to be positive.

That is we want:

`√(y-3/4)=x-1/2`

Add 1/2 on both sides:

`1/2+√(y-3/4)=x`

The last step is to switch x and y:

`1/2+√(x-3/4)=y`

`y=1/2+√(x-3/4)`

`f^(-1)(x)=1/2+√(x-3/4)`