Question

f:[1/2,infinity)—>[3/4,infinity)
f(x)=x^2-x+1
find the inverse of f(x);please explain

Answer 1

Answer:

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

Step-by-step 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:

$\pm \sqrt{y-3/4}=x-1/2$

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:

$\sqrt{y-3/4}=x-1/2$

Add 1/2 on both sides:

$1/2+\sqrt{y-3/4}=x$

The last step is to switch x and y:

$1/2+\sqrt{x-3/4}=y$

$y=1/2+\sqrt{x-3/4}$

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