Question

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = the quantity x minus seven divided by the quantity x plus three. and g(x) = quantity negative three x minus seven divided by quantity x minus one.

Answer 1

Answer: The prove is mentioned below.

Step-by-step explanation:

Here the given functions are,

$f(x)=\frac{x-7}{x+3}$

$g(x)=\frac{-3x-7}{x-1}$

We have to prove that: f(g(x)) = x and g(f(x)) = x.

Since $f(g(x))= \frac{(-3x-7)/(x-1) -1}{(-3x-7)/(x-1) +3}$

⇒$f(g(x))= \frac{-3x-7-7x+7}{-3x-7+3x-3}$

⇒$f(g(x))= \frac{-4x}{-4}$

⇒$f(g(x))= x$

Now, $g(f(x))= \frac{-3(x-7)/(x+3) -7}{(x-7)/(x+3) -1}$

⇒$g(f(x))= \frac{-3x+21-7x-21}{x-7-x-3}$

⇒$g(f(x))= \frac{-10x}{-10}$

⇒$g(f(x))= x$

Answer 2

Answer with Step-by-step explanation:

We are given that

$f(x)=\frac{x-7}{x+3}$

$g(x)=\frac{-3x-7}{x-1}$

We have to confirm that f and g are inverses by showing that f(g(x))=x=g(f(x))

$f(g(x))=f(\frac{-3x-7}{x-1})$

$f(g(x))=\frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3}$

$f(g(x))=\frac{\frac{-3x-7-7x+7}{x-1}}{\frac{-3x-7+3x-3}{x-1}}$

$f(g(x))=\frac{-10x}{x-1}\times \frac{x-1}{-10}$

$f(g(x))=x$

$g(f(x))=g(\frac{x-7}{x+3})$

$g(f(x))=\frac{-3(\frac{x-7}{x+3})-7}{\frac{x-7}{x+3}-1}$

$g(f(x))=\frac{\frac{-3x+21-7x-21}{x+3}}{\frac{x-7-x-3}{x+3}}$

$g(f(x))=\frac{\frac{-10x}{x+3}}{\frac{-10}{x+3}}$

$g(f(x))=\frac{-10x}{x+3}\times \frac{x+3}{-10}=x$

Hence, f(g(x))=g(f(x))=x

Therefore, f and g are inverses.