show that f(x) = x^2-1 (x is greater than and equal to 0) and g(x) = root(x-1) (x is greater than and equal to -1) are inverse functions.

when i try f(g(x)) = x and g(f(x)) = x to show that the two functions are inverses it doesn't work because i am not left with x when i solve the equation....

much help is needed thansk

They must not be inverse functions.

f(x)=x^2-1
y+1=x^2
x= +- root (y+1)
g(x)=+-root (x+1)
lets try the + root
f(g)=(x+1)-1=x
g(f)=+root (X^2-1+1)=x

where did you get the (x+1) from??

To show that two functions are inverse functions, you need to prove that when you compose them in both orders, you get back the original input. In other words, you need to show that f(g(x)) = x and g(f(x)) = x for all x in their domain.

Let's start by calculating f(g(x)):

f(g(x)) = f(sqrt(x-1))

To simplify this expression, we substitute g(x) = sqrt(x-1) into f:

f(g(x)) = f(sqrt(x-1)) = (sqrt(x-1))^2 - 1 = (x-1) - 1 = x - 2

Now let's calculate g(f(x)):

g(f(x)) = g(x^2 - 1)

To simplify this expression, we substitute f(x) = x^2 - 1 into g:

g(f(x)) = g(x^2 - 1) = sqrt((x^2 - 1) - 1) = sqrt(x^2 - 2)

Now, we have:

f(g(x)) = x - 2
g(f(x)) = sqrt(x^2 - 2)

From these expressions, we can see that f(g(x)) ≠ x and g(f(x)) ≠ x, which means that these two functions are not inverses of each other.

It is possible that there was a mistake in the initial problem or the descriptions of the functions provided. Please double-check the definitions of the functions f(x) and g(x) to ensure accuracy.