Let's solve each of these problems by finding $g(x)$ and $h(x)$ such that $f(x) = g(h(x))$:

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35(a) $f(x) = \sqrt{x+2}$

We want $f(x) = g(h(x))$. Let:

• $h(x) = x+2$
• $g(x) = \sqrt{x}$

Thus, $g(h(x)) = \sqrt{h(x)} = \sqrt{x+2}$.

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35(b) $f(x) = |x^2 - 3x + 5|$

We want $f(x) = g(h(x))$. Let:

• $h(x) = x^2 - 3x + 5$
• $g(x) = |x|$

Thus, $g(h(x)) = |h(x)| = |x^2 - 3x + 5|$.

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36(a) $f(x) = x^2 + 1$

We want $f(x) = g(h(x))$. Let:

• $h(x) = x^2$
• $g(x) = x + 1$

Thus, $g(h(x)) = h(x) + 1 = x^2 + 1$.

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36(b) $f(x) = \frac{1}{x-3}$

We want $f(x) = g(h(x))$. Let:

• $h(x) = x-3$
• $g(x) = \frac{1}{x}$

Thus, $g(h(x)) = \frac{1}{h(x)} = \frac{1}{x-3}$.

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37(a) $f(x) = \sin^2 x$

We want $f(x) = g(h(x))$. Let:

• $h(x) = \sin(x)$
• $g(x) = x^2$

Thus, $g(h(x)) = h(x)^2 = \sin^2 x$.

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37(b) $f(x) = \frac{3}{5+\cos x}$

We want $f(x) = g(h(x))$. Let:

• $h(x) = 5 + \cos(x)$
• $g(x) = \frac{3}{x}$

Thus, $g(h(x)) = \frac{3}{h(x)} = \frac{3}{5+\cos x}$.

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Would you like further clarifications or additional examples? Here are five related questions to consider:

1. Can you generalize the process of finding $g(x)$ and $h(x)$ for any given function $f(x)$?
2. What alternative choices for $g(x)$ and $h(x)$ could work for $f(x) = \sqrt{x+2}$?
3. How do we handle cases where $f(x)$ includes composite trigonometric functions?
4. What happens if $f(x)$ involves an absolute value with nested operations?
5. Can you verify these solutions by explicitly substituting $h(x)$ into $g(x)$?

Tip: Always choose $h(x)$ to simplify the innermost expression in $f(x)$, then define $g(x)$ accordingly.