Doubts in functions

Question

f(x) = 2x -1 , g(x) = 1/x, x not equal to 0.

g)x) = 1/x, x not equal to 0, h(x) = 2^x

Simplify hh^-1 (x)

Answer 1

Greetings! Lets solve this shall we ?

We have the function h(x) = 2^x. We must find h[h^-1(x)], correct? First, we must find the inverse of h(x), then plug it into the original function.

Then,

h^-1(x) = x = 2^y..................Then we take the log of both sides to obtain

log x = log 2^y....................Then we use the power rule of logarithms to bring down the y^-1 such that

log x = y^-1log2.............Finally, we divide log 2 to both sides to obtain the inverse

h^-1(x) = y^-1 = (logx/log2) ..............Now lets plug this into the original function such that

h[h^-1(x)] = 2^[logx/log2] ....................Now we must simplify this equation such that

h[h^-1(x)] = (2^(logx))^(log2)-1................. We swap the exponents using the algebra rule such that (xⁿ)^m = (x^m)ⁿ

h[h^-1(x)] = (2^(log2)-1)^logx ..........2^log2 becomes 1. Finally,

h[h^-1(x)] = (-1)^logx

I hope this helped!