Question 186434
*[Tex \LARGE f(x)=3x+2] Start with the given function.



*[Tex \LARGE y=3x+2] Replace f(x) with 'y'



*[Tex \LARGE x=3y+2] Switch x and y



*[Tex \LARGE x-2=3y] Subtract 2 from both sides.



*[Tex \LARGE \frac{x-2}{3}=y] Divide both sides by 3



*[Tex \LARGE y=\frac{x-2}{3}] Rearrange the equation.




So the inverse function is *[Tex \LARGE f^{-1}(x)=\frac{x-2}{3}] which is a function.





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*[Tex \LARGE f(x)=x^2-5] Start with the given function.



*[Tex \LARGE y=x^2-5] Replace f(x) with 'y'



*[Tex \LARGE x=y^2-5] Switch x and y



*[Tex \LARGE x+5=y^2] Add 5 to both sides.



*[Tex \LARGE y^2=x+5] Rearrange the equation



*[Tex \LARGE y=\pm \sqrt{x+5}] Take the square root of both sides. Note: don't forget the "plus/minus"



*[Tex \LARGE y=\sqrt{x+5}] or *[Tex \LARGE y=-\sqrt{x+5}] Break up the "plus/minus" to form two separate equations.



Since we have two separate equations, this means that we CANNOT write the inverse as one function.



So the inverse is NOT a function.