inverse composite function

**extraordinarymachine** · December 8th 2009, 05:10 PM

hi i need help on how to solve these questions

verify, algebraically, that $f(f{^-1}(x))$ = x

f(x) = $x^2$ -4

and

f(x) = $\frac{1}{x-2}$

**RRH** · December 8th 2009, 05:42 PM

Originally Posted by extraordinarymachine

hi i need help on how to solve these questions

verify, algebraically, that $f(f{^-1}(x))$ = x

f(x) = $x^2$ -4

and

f(x) = $\frac{1}{x-2}$

first you need to find the inverse of $x^2-4$

$y=x^2-4$

$x=y^2-4$

$x+4=y^2$

$\sqrt [2] {x+4}=\sqrt [2] {y^2}$

$\sqrt [2] {x}+2=y$

now we plug the inverse back into the original function

$(\sqrt [2] {x}+2)^2-4$

$x+4-4=x$

Can you follow the example to solve the second function

**HallsofIvy** · December 9th 2009, 04:11 AM

Originally Posted by extraordinarymachine

hi i need help on how to solve these questions

verify, algebraically, that $f(f{^-1}(x))$ = x

f(x) = $x^2$ -4

Unless you have some restriction on the possible values of x (on the domain) this function does not have an inverse!
RRH gets to $y^2= x+ 4$ and then says that $y= \sqrt{x+ 4}$ but there are two values of y whose square is x+ 4.

If the original function is " $f(x)= x^2- 4$ with domain $x\ge 0$ ", THEN $f^{-1}(x)= \sqrt{x+ 4}$ with domain $x\ge -2$ .

If the original function is " $f(x)= x^2- 4$ with domain $x\le 0$ , THEN $f^{-1}(x)= -\sqrt{x+ 4}$ with domain $x\ge -2$

RRH is also wrong when he writes $\sqrt{x+ 4}= \sqrt{x}+ 2$ . For example, if x= 1, $\sqrt{1+ 4}= \sqrt{5}$ which is certainly not equal to $\sqrt{1}+ 2= 3$ .

and

f(x) = $\frac{1}{x-2}$

Follow the pattern RRH used (which is correct). From $y= \frac{1}{x-2}$ (which has natural domain all real numbers except 2), the inverse function is given by $x= \frac{1}{y- 2}$ , where I have swapped x and y. Now solve for y.

**RRH** · December 9th 2009, 04:41 AM

Originally Posted by HallsofIvy

Unless you have some restriction on the possible values of x (on the domain) this function does not have an inverse!
RRH gets to $y^2= x+ 4$ and then says that $y= \sqrt{x+ 4}$ but there are two values of y whose square is x+ 4.

If the original function is " $f(x)= x^2- 4$ with domain $x\ge 0$ ", THEN $f^{-1}(x)= \sqrt{x+ 4}$ with domain $x\ge -2$ .

If the original function is " $f(x)= x^2- 4$ with domain $x\le 0$ , THEN $f^{-1}(x)= -\sqrt{x+ 4}$ with domain $x\ge -2$

RRH is also wrong when he writes $\sqrt{x+ 4}= \sqrt{x}+ 2$ . For example, if x= 1, $\sqrt{1+ 4}= \sqrt{5}$ which is certainly not equal to $\sqrt{1}+ 2= 3$ .

Follow the pattern RRH used (which is correct). From $y= \frac{1}{x-2}$ (which has natural domain all real numbers except 2), the inverse function is given by $x= \frac{1}{y- 2}$ , where I have swapped x and y. Now solve for y.

thank you for the correction I was thinking about that last night ... I have been sloppy with my math lately ... Again I apologize to the original poster to the original poster and to others in the MHF for my mistakes

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