Functions - Composition of functions

Given $f(x) = 2x + 5$ and $g(x) = x^2 - 3$, find $(f \circ g)(x)$ and $(g \circ f)(x)$.

$(f \circ g)(x) = f(g(x)) = f(x^2 - 3) = 2(x^2 - 3) + 5 = 2x^2 - 6 + 5 = 2x^2 - 1$ $(g \circ f)(x) = g(f(x)) = g(2x + 5) = (2x + 5)^2 - 3 = 4x^2 + 20x + 25 - 3 = 4x^2 + 20x + 22$

To find $f(g(x))$, substitute the entire expression for $g(x)$ into every $x$ in $f(x)$. To find $g(f(x))$, substitute the expression for $f(x)$ into every $x$ in $g(x)$.

Let $h(x) = \sqrt{x-1}$ and $k(x) = \frac{1}{x}$. Determine the expression for $(k \circ h)(x)$ and state its domain.

$(k \circ h)(x) = k(h(x)) = k(\sqrt{x-1}) = \frac{1}{\sqrt{x-1}}$ Domain: For $\sqrt{x-1}$ to be defined, $x - 1 \ge 0 \implies x \ge 1$. For the fraction to be defined, the denominator cannot be zero, so $\sqrt{x-1} \neq 0 \implies x \neq 1$. Combining these, the domain is $x > 1$.

The domain is restricted by the inner function (which requires $x \ge 1$) and the outer function (which prevents the denominator from being zero at $x=1$).

If $f(x) = 3x - 2$, find the value of $x$ such that $(f \circ f)(x) = 19$.

First, find $(f \circ f)(x)$: $(f \circ f)(x) = f(3x - 2) = 3(3x - 2) - 2 = 9x - 6 - 2 = 9x - 8$ Set the expression equal to $19$: $9x - 8 = 19$ $9x = 27$ $x = 3$

Find the composite function $f(f(x))$ first by substituting the function into itself, then solve the resulting linear equation for $x$.

1. Write the composite expression: $(f \circ g)(x) = f(g(x))$
2. Substitute $g(x)$ into $f$: $f(x+3) = (x+3)^2$
3. Expand the expression (optional): $h(x) = x^2 + 6x + 9$
4. Identify the vertex: Since the function is in the form $(x-h)^2 + k$, where $h = -3$ and $k = 0$, the vertex is at $(-3, 0)$.

This example demonstrates how a horizontal translation $g(x)$ shifts the base function $f(x) = x^2$ to the left by 3 units before the squaring operation occurs.

1. Express the composite function: $(f \circ g)(x) = f(2x - 4) = \frac{1}{2x - 4}$.
2. Determine the restriction: The denominator of a fraction cannot be zero.
3. Solve $2x - 4 = 0$ which gives $2x = 4 \implies x = 2$.
4. State the domain: $x \in \mathbb{R}, x \neq 2$.

The domain of a composite function must exclude values that make the inner function undefined (none here) and values where the output of the inner function is not in the domain of the outer function (where $g(x) = 0$).