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Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

f(x) = quantity x minus seven divided by quantity x plus two. and g(x) = quantity negative two x minus seven divided by quantity x minus one.

Respuesta :

nobillionaireNobley
f(x) = (x - 7)/(x + 2)
g(x) = (-2x - 7)/(x - 1)
f(g(x)) = [(-2x - 7)/(x-1) - 7]/[(-2x - 7)/(x-1) + 2]
= [(-2x - 7 - 7x + 7)/(x - 1)]/[(-2x - 7 + 2x - 2)/(x - 1)]
=(-9x)/-9
= x

g(x) = (-2x - 7)/(x - 1)
g(f(x)) = [-2(x - 7)/(x + 2) - 7]/[(x - 7)/(x + 2) - 1]
= [(-2x + 14 - 7x - 14)/(x + 2)]/[(x - 7 - x - 2)/(x + 2)]
= -9x/-9
= x

i.e. f(g(x)) = g(f(x)) = x
which shows that they are inverses.
tifanihayyu

F and g are inverses by showing that f(g(x)) = x and g(f(x)) = x

Further explanation

f(x) = quantity x minus seven divided by quantity x plus two. It means we should write it as follows:

[tex]f(x) = \frac{x-7}{x+2}[/tex]

And g(x) = quantity negative two x minus seven divided by quantity x minus one.  It means we should write it as follows:

[tex]g(x)=\frac{-2x-7}{x-1}[/tex]

F and g are inverses by showing that f(g(x)) = x and g(f(x)) = x

[tex]f(g(x)) = x \\f(g(x)) = \frac{\frac{-2x - 7}{(x-1) - 7} }{ \frac{-2x - 7}{(x-1) + 2} } \\f(g(x)) = \frac{ \frac{-2x - 7 - 7x + 7}{(x - 1)} }{ \frac{-2x - 7 + 2x - 2}{(x - 1)} } \\f(g(x)) = \frac{-9x}{-9} = x[/tex]

[tex]g(x) = \frac{-2x - 7}{x - 1} \\g(x) = \frac{\frac{-2(x - 7)}{(x + 2) - 7}}{\frac{(x - 7)}{(x + 2) - 1} } \\g(x) = \frac{\frac{(-2x + 14 - 7x - 14)}{(x + 2)} }{\frac{(x - 7 - x - 2)}{x + 2} } \\g(x) = \frac{-9x}{-9} = x[/tex]

Therefore:

It is confirmed that f and g are inverses by showing that [tex]f(g(x)) = g(f(x)) = x[/tex].

Learn more

  1. Learn more about inverse https://brainly.com/question/2541698
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  3. Learn more   about minus https://brainly.com/question/5033632

Answer details

Grade:    5

Subject:  Math

Chapter:  inverses

Keywords: inverse, quantity, minus, divided, negative