Question

find two functions f and g such that (f ∘ g)(x) = h(x). (there are many correct answers. use non-identity functions for f(x) and g(x))
h(x) = \sqrt{9 - x}
(f(x), g(x)) = \left(\\ \
ight)

Explanation:

Step1: Analyze the composite function

We know that \((f \circ g)(x)=f(g(x)) = h(x)=\sqrt{9 - x}\). We need to find \(f(x)\) and \(g(x)\) such that when we substitute \(g(x)\) into \(f(x)\), we get \(\sqrt{9 - x}\). A common way is to look at the "inner" function and the "outer" function. The inner function can be \(g(x)=9 - x\) and the outer function can be \(f(x)=\sqrt{x}\) because if we substitute \(g(x)\) into \(f(x)\), we get \(f(g(x))=\sqrt{g(x)}=\sqrt{9 - x}\), which is equal to \(h(x)\).

Step2: Verify the composition

Let's check the composition. If \(f(x)=\sqrt{x}\) and \(g(x) = 9 - x\), then \((f\circ g)(x)=f(g(x))=f(9 - x)=\sqrt{9 - x}\), which is exactly \(h(x)\).

Answer:

\(f(x)=\sqrt{x}\), \(g(x)=9 - x\) (There are other possible answers, for example, \(f(x)=\sqrt{9 - x}\) and \(g(x)=x\) is not allowed as we need non - identity functions for \(f\) and \(g\) in the sense of not being trivial, but the above is a valid non - trivial pair)