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Georgia [21]
3 years ago
8

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

Mathematics
1 answer:
dexar [7]3 years ago
8 0

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}

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