Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The original function is $f(x)= (3x-2)^.5$

find $y=f^{-1}(x)$ and its domain.

So I found the inverse equation to be $y=((x^2)+2)/3 $ The correct answer for the domain is all reals when $x \geq 0$. Why is this so? Why would it not be all real numbers?

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

The domain of $f^{-1}$ is the range of $f$. Your function is $f(x)=\sqrt{3x-2}$, $x\ge \frac23$. Obviously, $f(x)\ge 0$ and so the range is at most the non-negative reals. Let $y\ge 0$. Then $$y=f(x)\iff y=\sqrt{3x-2}\iff y^2=3x-2\iff x=\frac{y^2+2}3$$ You must check whether or not $x\ge \frac23$. $$x\ge \frac23\iff \frac{y^2+2}3\ge \frac23\iff y^2\ge 0$$ which is true. Thus the domain of $f^{-1}$ (I suppose you have already proven $f$ is 1-1) is $[0,+\infty)$ and $$f^{-1}(x)=\frac{x^2+2}3$$

share|improve this answer
    
+1 for sportsmanship, (Not the badge) –  Rustyn Jan 26 '13 at 7:25
    
@user59714: Note that on $I:[2/3,\infty)$, the function is increasing strictly and so it has an inverse. +1 for straightforward answer. –  B. S. Jan 26 '13 at 7:53
add comment

The domain of the inverse function, $f^{-1}(x)$, takes on the range of the function, $f(x)$. Your original function's range is the set of non-negative real numbers, so hence the set of non-negative real numbers must be the domain of $f^{-1}(x)$.

share|improve this answer
    
+1 Nice approach. ;-) –  B. S. Jan 26 '13 at 7:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.