Question: Suppose that the functions $h$ and $f$ are defined as follows.
$h(x) = x^2 + 3$
$f(x) = \frac{5}{3x}, x \neq 0$
Find the compositions $h \circ h$ and $f \circ f$.
Simplify your answers as much as possible.
(Assume that your expressions are defined for all $x$ in the domain of the composition. You do not have to indicate the domain.)
$(h \circ h)(x) =$
$(f \circ f)(x) =$
1. **State the problem:** We are given two functions:
$$h(x) = x^2 + 3$$
$$f(x) = \frac{5}{3x}, x \neq 0$$
We need to find the compositions $(h \circ h)(x)$ and $(f \circ f)(x)$ and simplify them.
2. **Recall the composition definition:**
For two functions $g$ and $k$, the composition $(g \circ k)(x) = g(k(x))$.
3. **Find $(h \circ h)(x)$:**
This means $h(h(x))$.
Substitute $h(x)$ into $h$:
$$h(h(x)) = h\left(x^2 + 3\right) = \left(x^2 + 3\right)^2 + 3$$
4. **Simplify $(x^2 + 3)^2 + 3$:**
Expand the square:
$$\left(x^2 + 3\right)^2 = (x^2)^2 + 2 \cdot x^2 \cdot 3 + 3^2 = x^4 + 6x^2 + 9$$
Add 3:
$$x^4 + 6x^2 + 9 + 3 = x^4 + 6x^2 + 12$$
5. **Therefore:**
$$(h \circ h)(x) = x^4 + 6x^2 + 12$$
6. **Find $(f \circ f)(x)$:**
This means $f(f(x))$.
Substitute $f(x)$ into $f$:
$$f(f(x)) = f\left(\frac{5}{3x}\right) = \frac{5}{3 \cdot \frac{5}{3x}}$$
7. **Simplify the denominator:**
$$3 \cdot \frac{5}{3x} = \frac{15}{3x} = \frac{15}{3x}$$
Actually, better to keep as $3 \times \frac{5}{3x} = \frac{15}{3x}$ is incorrect, let's do carefully:
$$3 \times \frac{5}{3x} = \frac{3 \times 5}{3x} = \frac{15}{3x}$$
Simplify numerator and denominator:
$$\frac{15}{3x} = \frac{15}{3x}$$
But better to write as:
$$\frac{5}{3 \times \frac{5}{3x}} = \frac{5}{\frac{15}{3x}}$$
8. **Dividing by a fraction:**
$$\frac{5}{\frac{15}{3x}} = 5 \times \frac{3x}{15} = \frac{15x}{15}$$
9. **Simplify the fraction:**
$$\frac{15x}{15} = \cancel{15}x / \cancel{15} = x$$
10. **Therefore:**
$$(f \circ f)(x) = x$$
**Final answers:**
$$(h \circ h)(x) = x^4 + 6x^2 + 12$$
$$(f \circ f)(x) = x$$