To prove that the inverse function of h(x) is h^−1(x) = f^−1(g^−1(x)), we need to show that applying h^−1 after h leaves us with the original value.

Let's start with the definition of h(x) = g(f(x)).

1. Apply h to some input x. This gives us h(x) = g(f(x)).
2. Now, apply the inverse h^−1 to h(x). According to the claim, this is equivalent to applying f^−1 to g^−1 to g(f(x)).

So, we have h^−1(h(x)) = h^−1(g(f(x))) = f^−1(g^−1(g(f(x)))).

Since g^−1(g(y)) = y for any y (because g is a one-to-one function and thus invertible), we can simplify f^−1(g^−1(g(f(x)))) to f^−1(f(x)).

Similarly, since f^−1(f(x)) = x (because f is also a one-to-one function and thus invertible), we can simplify f^−1(f(x)) to x.

So, we have shown that h^−1(h(x)) = x, which means that h^−1(x) = f^−1(g^−1(x)) is indeed the inverse function of h(x).

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To prove that the inverse function of h(x) is h^−1(x) = f^−1(g^−1(x)), we need to show that applying h^−1 after h leaves us with the original value.

Let's start with the definition of h(x) = g(f(x)).

1. Apply h to some input x. This gives us h(x) = g(f(x)).
2. Now, apply the inverse h^−1 to h(x). According to the claim, this is equivalent to applying f^−1 to g^−1 to g(f(x)).

So, we have h^−1(h(x)) = h^−1(g(f(x))) = f^−1(g^−1(g(f(x)))).

Since g^−1(g(y)) = y for any y (because g is a one-to-one function and thus invertible), we can simplify f^−1(g^−1(g(f(x)))) to f^−1(f(x)).

Similarly, since f^−1(f(x)) = x (because f is also a one-to-one function and thus invertible), we can simplify f^−1(f(x)) to x.

So, we have shown that h^−1(h(x)) = x, which means that h^−1(x) = f^−1(g^−1(x)) is indeed the inverse function of h(x).

Knowee AI · Accepted Answer

To prove that the inverse function of h(x) is h^−1(x) = f^−1(g^−1(x)), we need to show that applying h^−1 after h leaves us with the original value.

Let's start with the definition of h(x) = g(f(x)).

1. Apply h to some input x. This gives us h(x) = g(f(x)).
2. Now, apply the inverse h^−1 to h(x). According to the claim, this is equivalent to applying f^−1 to g^−1 to g(f(x)).

So, we have h^−1(h(x)) = h^−1(g(f(x))) = f^−1(g^−1(g(f(x)))).

Since g^−1(g(y)) = y for any y (because g is a one-to-one function and thus invertible), we can simplify f^−1(g^−1(g(f(x)))) to f^−1(f(x)).

Similarly, since f^−1(f(x)) = x (because f is also a one-to-one function and thus invertible), we can simplify f^−1(f(x)) to x.

So, we have shown that h^−1(h(x)) = x, which means that h^−1(x) = f^−1(g^−1(x)) is indeed the inverse function of h(x).

Let f(x) and g (x) be one-to-one functions, and let h (x) = g ( f(x)). Show that the inverse functionof h (x) is h^−1 (x) = f^ −1(g^−1 (x))

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