Question 323243
To check if a function is the inverse of another function, we use the property *[Tex \LARGE f(f^{-1}(x))=x] and we also use *[Tex \LARGE f^{-1}(f(x))=x]. We have to check both equations. 



Since we're given that *[Tex \LARGE f(x)=4x-8] and you claim that *[Tex \LARGE f^{-1}(x)=x+2], this means that *[Tex \LARGE f(f^{-1}(x))=4(x+2)-8=4x+8-8=4x] which is NOT equal to 'x'. So *[Tex \LARGE f^{-1}(x)=x+2] is not correct.


---------------------------------------------------------


Let's find the inverse of *[Tex \LARGE f(x)=4x-8]



*[Tex \LARGE f(x)=4x-8] Start with the given function.



{{{y=4x-8}}} Replace f(x) with y



{{{x=4y-8}}} Swap x and y. The goal now is to solve for y



{{{x+8=4y}}} Add 8 to both sides.



{{{(x+8)/4=y}}} Divide both sides by 4 to isolate y.



{{{y=(x+8)/4}}} Rearrange the equation



{{{y=x/4+8/4}}} Break up the fraction



{{{y=(1/4)x+2}}} Reduce and simplify.



So the inverse function of *[Tex \LARGE f(x)=4x-8] is *[Tex \LARGE f^{-1}(x)=\frac{1}{4}x+2]


---------------------------

Check:


Let's first see if *[Tex \LARGE f(f^{-1}(x))=x]



*[Tex \LARGE f(x)=4x-8] ... Start with the given function.



*[Tex \LARGE f(f^{-1}(x))=4\left(\frac{1}{4}x+2\right)-8] ... Plug in *[Tex \LARGE f^{-1}(x)=\frac{1}{4}x+2]



*[Tex \LARGE f(f^{-1}(x))=4\left(\frac{1}{4}x\right)+4\left(2\right)-8] ... Distribute.



*[Tex \LARGE f(f^{-1}(x))=\frac{4}{4}x+8-8] ... Multiply



*[Tex \LARGE f(f^{-1}(x))=\frac{4}{4}x] ... Combine like terms.



*[Tex \LARGE f(f^{-1}(x))=x] ... Reduce.



Now let's see if *[Tex \LARGE f^{-1}(f(x))=x] 



*[Tex \LARGE f^{-1}(x)=\frac{1}{4}x+2] ... Start with the inverse function.



*[Tex \LARGE f^{-1}(f(x))=\frac{1}{4}(4x-8)+2] ... Plug in *[Tex \LARGE f(x)=4x-8]



*[Tex \LARGE f^{-1}(f(x))=\frac{1}{4}(4x)+\frac{1}{4}(-8)+2] ... Distribute.



*[Tex \LARGE f^{-1}(f(x))=\frac{4x}{4}+\frac{-8}{4}+2] ... Multiply



*[Tex \LARGE f^{-1}(f(x))=x-2+2] ... Reduce



*[Tex \LARGE f^{-1}(f(x))=x] ... Combine like terms.



So we've shown that *[Tex \LARGE f(f^{-1}(x))=x] and *[Tex \LARGE f^{-1}(f(x))=x]



So this verifies that the inverse function of *[Tex \LARGE f(x)=4x-8] is indeed *[Tex \LARGE f^{-1}(x)=\frac{1}{4}x+2]