The image contains function composition problems. Let's solve each question step by step.

First Problem

Given:

• $f(x) = x^2$
• $g(x) = \sqrt{x - 1}$

Questions:

1. $(f \circ g)(x)$
2. $(g \circ f)(x)$

Solutions:

1. $(f \circ g)(x)$:

   • Composition means applying $g(x)$ first and then $f(x)$.
   • $(f \circ g)(x) = f(g(x))$.
   • Substitute $g(x)$: $f(g(x)) = f(\sqrt{x - 1})$.
   • Now apply $f(x) = x^2$: $f(\sqrt{x - 1}) = (\sqrt{x - 1})^2 = x - 1$.

   So, $(f \circ g)(x) = x - 1$.

2. $(g \circ f)(x)$:

   • Composition means applying $f(x)$ first and then $g(x)$.
   • $(g \circ f)(x) = g(f(x))$.
   • Substitute $f(x)$: $g(f(x)) = g(x^2)$.
   • Now apply $g(x) = \sqrt{x - 1}$: $g(x^2) = \sqrt{x^2 - 1}$.

   So, $(g \circ f)(x) = \sqrt{x^2 - 1}$.

Second Problem

Given:

• $f(x) = x^2 - 3x$
• $g(x) = 4 - \sqrt{x} + 2$
• $h(x) = \frac{2x}{x - 1}$

Questions:

1. $(f \circ g)(x)$
2. $(g \circ f)(x)$
3. $(g \circ h)(x)$
4. $(h \circ (g \circ f))(x)$

Solutions:

1. $(f \circ g)(x)$:

   • $(f \circ g)(x) = f(g(x))$.
   • Substitute $g(x)$: $f(g(x)) = f(4 - \sqrt{x} + 2) = f(6 - \sqrt{x})$.
   • Now apply $f(x) = x^2 - 3x$: $f(6 - \sqrt{x}) = (6 - \sqrt{x})^2 - 3(6 - \sqrt{x}).$
   • Expand and simplify: $(6 - \sqrt{x})^2 = 36 - 12\sqrt{x} + x,$ and $-3(6 - \sqrt{x}) = -18 + 3\sqrt{x}.$ So, $f(6 - \sqrt{x}) = 36 - 12\sqrt{x} + x - 18 + 3\sqrt{x} = x - 9\sqrt{x} + 18.$

   So, $(f \circ g)(x) = x - 9\sqrt{x} + 18$.

2. $(g \circ f)(x)$:

   • $(g \circ f)(x) = g(f(x))$.
   • Substitute $f(x)$: $g(f(x)) = g(x^2 - 3x)$.
   • Now apply $g(x) = 4 - \sqrt{x} + 2$: $g(x^2 - 3x) = 4 - \sqrt{x^2 - 3x} + 2 = 6 - \sqrt{x^2 - 3x}.$

   So, $(g \circ f)(x) = 6 - \sqrt{x^2 - 3x}$.

3. $(g \circ h)(x)$:

   • $(g \circ h)(x) = g(h(x))$.
   • Substitute $h(x)$: $g(h(x)) = g\left(\frac{2x}{x - 1}\right)$.
   • Apply $g(x) = 4 - \sqrt{x} + 2$: $g\left(\frac{2x}{x - 1}\right) = 6 - \sqrt{\frac{2x}{x - 1}}.$

   So, $(g \circ h)(x) = 6 - \sqrt{\frac{2x}{x - 1}}$.

4. $(h \circ (g \circ f))(x)$:

   • First find $(g \circ f)(x)$: $6 - \sqrt{x^2 - 3x}$.
   • Now, $h((g \circ f)(x)) = h(6 - \sqrt{x^2 - 3x})$.
   • Apply $h(x) = \frac{2x}{x - 1}$: $h(6 - \sqrt{x^2 - 3x}) = \frac{2(6 - \sqrt{x^2 - 3x})}{(6 - \sqrt{x^2 - 3x}) - 1}.$

   So, $(h \circ (g \circ f))(x) = \frac{2(6 - \sqrt{x^2 - 3x})}{(6 - \sqrt{x^2 - 3x}) - 1}$.

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Related Questions

1. How do you simplify compositions of functions involving radicals?
2. What are the properties of function compositions?
3. How do you handle domain restrictions in function compositions?
4. How to graph the compositions of these functions?
5. How can compositions be used in real-world applications?

Tip: Always check the domain of each function and the composed function to ensure they are valid for all given values.