Question

Confirm that f and g are inverses: f(x)=(x+1)/x and g(x)=1/(x-1)

Answer 1

to confirm that that is the case, you'd do a f( g(x) ) , and that should equals to "x"
and g( f(x) ) and that should also equals to "x"

lemme do the 1st case
\(\bf f(x)=\cfrac{x+1}{x}\qquad \qquad g(x)=\cfrac{1}{x-1}\\ \quad \\ \quad \\
f(\quad g(x)\quad ) = \cfrac{\left(\frac{1}{x-1}\right)+1}{\left(\frac{1}{x-1}\right)}\\ \quad \\

\cfrac{\left(\frac{1}{x-1}\right)+1}{\left(\frac{1}{x-1}\right)}\implies \cfrac{\frac{1}{x-1}+\frac{1}{1}}{\left(\frac{1}{x-1}\right)}\implies
\cfrac{\frac{\cancel{1}+(x\cancel{-1})}{x-1}}{\left(\frac{1}{x-1}\right)}\\ \quad \\

\cfrac{\frac{x}{x-1}}{\frac{1}{x-1}}\implies \cfrac{x}{\cancel{x-1}}\times \cfrac{\cancel{x-1}}{1}\implies x\)

Answer 2

now try the 2nd one :)

Answer 3

how did become ?

Answer 4

ohh the addition? well the LCD is x-1, if you use that as the LCD, see what you'd get as the numerator

Answer 5

can you do the 2nd one?

Answer 6

g( f(x) )

Answer 7

\(\bf f(x)=\color{blue}{\cfrac{x+1}{x}}\qquad \qquad g(x)=\cfrac{1}{x-1}\\ \quad \\ \quad \\
g(\quad f(x)\quad ) = g(x)=\cfrac{1}{\left(\color{blue}{\frac{x+1}{x}}\right)-1}\)

Answer 8

the 1st one gave 'x" as result, so they're indeed inverse of each other, so the 2nd will also give "x", try it

Answer 9

keep getting stuck at

Answer 10

well... if the LCD is "x" at the bottom,\(\bf x \div x = 1\), so the 1st term is (x+1)
the 2nd term will be \(\bf x\div 1 \ne 1\)

Answer 11

\(\bf g(\quad f(x)\quad ) = g(x)=\cfrac{1}{\left(\frac{x+1}{x}\right)-1}\\ \quad \\
\cfrac{1}{\frac{x+1}{x}-1}\implies \cfrac{1}{\frac{(x+1)-x}{x}}\)

Answer 12

so

Answer 13

thank you

Answer 14

yeap \(\bf \cfrac{1}{\frac{x+1}{x}-1}\implies \cfrac{1}{\frac{(\cancel{x}+1)\cancel{-x}}{x}}\implies \cfrac{1}{\frac{1}{x}}\implies \cfrac{1}{1}\times \cfrac{x}{1}\implies x\)

Answer 15

yw