Function notation

[kk_kk]

I was given the funtion "f(x)=3x2+10" and was told to solve f^-1(x)[the inverse]. i got "f^-1(x)=sqrt( (x-10)/3 )" i would really appriciate it if someone could check my work. They also told to slove "f^-1( f(3) )" and im not sure what i need to do.[edited]

 

[Deleted member 4993]

I was given the funtion "f(x)=3x2+10" and was told to solve f^-1(x)[the inverse]. i got "f^-1(x)=sqrt( (x-10)/3 )" i would really appriciate it if someone could check my work. They also told to slove "f^-1( f(3) )" and im not sure what i need to do.[edited]

f(3) = 3*9+10 = 37

\(\displaystyle f^{-1}(37) \ = \sqrt{\frac{37 - 10}{3}} \ = \ ? \)

 

[HallsofIvy]

If y= 3x^2+ 10 (please don't write "3x2", that's very confusing!) then the inverse is given by x= 3y^2+ 10. Solve for y: x- 10= 3y^2. y^2= (x- 10)/3. Now, here's a problem. The original function, 3x^2+ 10 does NOT actually have an inverse! That's because the function is not "one to one". Both x= 1 and x= -1 give the same y value, 13. The inverse function has to assign to x= 13 both y= 1 and y= -1. But a function can't do that.

What has been done here is to choose one. At y^2= (x-10)/3 we have to take the square root so apparently the positive root has been (arbitrarily) chosen- y= sqrt((x- 10)/3).

Now you have been asked to find (not solve and certainly not "slove" ?)\(\displaystyle f^{-1}(f(3))\).

\(\displaystyle f(x)= 3x^2+ 10\) so \(\displaystyle f(3)= 3(3^2)+ 10= 3(9)+ 10= 27+ 10= 37\).

Now \(\displaystyle f^{-1}(f(3))= f^{-1}(37)= \sqrt{\frac{37- 10}{3}}= \sqrt{\frac{27}{3}}=\sqrt{9}= 3\).

That should be no surprize!

 

[Steven G]

To be sure that you found the inverse you should compute f(f^-1)(x) and verify that you get back x.