Question

Find functions \(f, g\), and h such that \(F=f \circ g \circ h .\) (Note: The answer is not unique.) a. \(F(x)=\frac{1}{\left(2 x^{2}+x+3\right)^{3}}\) b. \(F(x)=\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\)

Short answer

In summary, for the given functions: a. \(F(x) = \frac{1}{\left(2x^2 + x + 3\right)^3}\), a possible decomposition into f, g, and h functions could be: - h(x) = \(2x^2 + x + 3\) - g(x) = \(x^{-3}\) - f(x) = x b. \(F(x) = \frac{\sqrt{x + 1} - 1}{\sqrt{x + 1} + 1}\), a possible decomposition into f, g, and h functions could be: - h(x) = \(x + 1\) - g(x) = \(\sqrt{x}\) - f(u) = \(\frac{u - 1}{u + 1}\)

Step-by-step solution

Identify a function for h

We can identify that the most inner part of the function is \(2x^2 + x + 3\). So, let's consider h(x) = \(2x^2 + x + 3\).

Identify a function for g

Now that we know h(x), we can see that the next operation is taking the power of -3. So, let's consider g(x) = \(x^{-3}\).

Identify a function for f

Since g(x) and h(x) together form the given function, we can have f(x) = x. In this decomposition, we have:\[ F(x) = f(g(h(x))) = f(g(2x^2 + x + 3)) = f((2x^2 + x + 3)^{-3}) = (2x^2 + x + 3)^{-3} \] b. \(F(x) = \frac{\sqrt{x + 1} - 1}{\sqrt{x + 1} + 1}\)

Identify a function for h

We can identify that the most inner part of the function is \(x + 1\). So, let's consider h(x) = \(x + 1\).

Identify a function for g

Now that we know h(x), we can see that taking the square root is the next operation. So, let's consider g(x) = \(\sqrt{x}\).

Identify a function for f

Now we have h(x) and g(x), so we can look for the f(x) function. Since g(h(x)) = \(\sqrt{x + 1}\), we can rewrite the given F(x) with respect to g(h(x)) as follows:\[ F(x) = \frac{g(h(x)) - 1}{g(h(x)) + 1} \] Therefore, we can set f(u) as, \[ f(u) = \frac{u - 1}{u + 1} \] In this decomposition, we have:\[ F(x) = f(g(h(x))) = f(g(x + 1)) = f(\sqrt{x + 1}) = \frac{\sqrt{x + 1} - 1}{\sqrt{x + 1} + 1} \]