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If f(x) and r^(x) are inverse functions of eachother and f(x)=2x+5 what is f-1(8)?

User Foto
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recall that for inverse functions, the range of the original is the domain of the inverse and othe other way around.

so f⁻¹(8) is f⁻¹(x) but making x = 8, so the domain value in this case for the inverse is 8, and that's going to give us something, but let's nevermind that.

since the domain value is 8 for the inverse, that means, the same 8, is a range for the original f(x).

meaning, "some value of x" on the original, gives us a range of 8, then let's use that 8 in the original, BUT not on "x", but on "y", or f(x).


\bf f^(-1)(8)~\hspace{7em}\stackrel{f(x)}{8}=2x+5 \\\\[-0.35em] ~\dotfill\\\\ 8=2x+5\implies 3=2x\implies \boxed{\cfrac{3}{2}=x} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{original f(x)}}{\left(\stackrel{x}{(3)/(2)}~~,~~\stackrel{y}{8} \right)}~\hspace{7em} \stackrel{\textit{inverse }f^(-1)(x)}{\left(\stackrel{x}{8}~~,~~\stackrel{y}{(3)/(2)} \right)}~\hfill f^(-1)(8)=\cfrac{3}{2}


so notice how the values swap places for the inverses, one's domain, is the other's range.


User Ori Price
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