# Determine the inverse g(x) of the function f(x)=1+1/x , stating its domain and range. Verify that f(g(x)) = g(f(x)) = x and that g’(f(x))= 1/(f’(x))

can anyone kindly show me how to do this question?

Any help is appreciated.

• What have you tried so far? – Weston Miller Apr 11 at 16:27
• Please show us what you have tried. Hint: write $f(x) = 1 + \frac{1}{x}$ as $y= 1 + \frac{1}{x}$ – Knight Apr 11 at 16:27

$$f:(0,\infty)\rightarrow (1,\infty), f(x)=1+1/x$$ $$y=1+1/x \implies x=\frac{1}{y-1} \implies f^{-1}(x)=\frac{1}{x-1}=g(x)$$ $$\implies f(g(x))=1+\frac{1}{g(x)}=1+\frac{1}{1/(x-1)}=1+x-1=x$$ Also, $$g(f(x))=g(1+1/x)=g(\frac{x+1}{x})=\frac{1}{(x+1)/x-1}=x$$ Next, $$g(f(x))=x \implies g'(f(x)) f'(x)=1 \implies g'(f(x))=\frac{1}{f'(x)}.$$

Another function is possible such that $$f: (-\infty, 0) \rightarrow(-\infty, 1)$$ and it can be handled accordingly as above.