Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - Review Exercises: 27

Answer

$f(x)=x^{2}-4x+5$ $f^{-1}(x)=2+\sqrt{x-1}$
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Work Step by Step

$f(x)=x^{2}-4x+5$, for $x\gt2$ Substitute $f(x)$ by $y$: $y=x^{2}-4x+5$ Group the first two terms on the right side of the equation together: $y=(x^{2}-4x)+5$ Complete the square for the expression inside parentheses. Do so by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside parentheses and subtracting it from the expression outside of the parentheses. In this case, $b=-4$ $y=\Big[x^{2}-4x+\Big(-\dfrac{4}{2}\Big)^{2}\Big]+5-\Big(-\dfrac{4}{2}\Big)^{2}$ $y=(x^{2}-4x+4)+5-4$ $y=(x^{2}-4x+4)+1$ Factor the expression inside parentheses, which is a perfect square trinomial: $y=(x-2)^{2}+1$ Take $1$ to the left side and rearrange: $y-1=(x-2)^{2}$ $(x-2)^{2}=y-1$ Take the square root of both sides: $\sqrt{(x-2)^{2}}=\sqrt{y-1}$ $x-2=\sqrt{y-1}$ Take $2$ to the right side: $x=2+\sqrt{y-1}$ Interchange $x$ and $y$: $y=2+\sqrt{x-1}$ Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=2+\sqrt{x-1}$ The graph of the function and its inverse is shown in the answer section.
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