First, make sure you have typed things correctly. What you have written is gibberish. So, I shall go with
g(x) = -4/(-x-3) + 1
First of all, what's with all those - signs? Oh, well ...
swap variables and solve for x:
h(x) = 4/(x-1) - 3 = (7-3x)/(x-1)
By definition, f^-1(f(x)) = f(f^-1(x)) = x
So, to show that g and h are inverses, just evaluate g(h) and h(g)
g(h(x)) = -4/(-h(x)-3) + 1
= -4/(-(7-3x)/(x-1)-3)+1
= 4/((7-3x)/(x-1)+3)+1
= 4/((7-3x+3x-3))/(x-1))+1
= 4/(4/(x-1))+1
= 4 * (x-1)/4 + 1
= x-1+1
= x
Work h(g(x)) the same way
After you fix up your functions with the proper use of parentheses, of course!
Prove if the given functions are inverses by evaluating the composition f(h(x)) or h(f(x))
g(x) = -(4/-x-3) +1
f(x)= (3/-x-1) -2
How do we go about once we for example plug in the first expression into x in the second expression, in terms of what to do exactly.
Thanks
4 answers
oh. sorry, what exactly happened at the “swap variables and solve for x” step? How did you get (7-3x)/(x-1)
Ooops my bad. If you swap first, solve for y
you need to review how to find an inverse function.
If y = 2x-3
to find f^-1, first swap variables
x = 2y-3
then solve for y
y = (x+3)/2
Or, you could solve for x and then swap
y = 2x-3
y-3 = 2x
(y-3)/2 = x
x = y^-1 = (x-3)/2
you need to review how to find an inverse function.
If y = 2x-3
to find f^-1, first swap variables
x = 2y-3
then solve for y
y = (x+3)/2
Or, you could solve for x and then swap
y = 2x-3
y-3 = 2x
(y-3)/2 = x
x = y^-1 = (x-3)/2
y+3