Functions - Composite and inverse functions (HL)

Given $f(x) = 2x + 5$ and $g(x) = x^2 - 3$, find the composite function $(g \circ f)(x)$ and evaluate $(g \circ f)(1)$.

$(g \circ f)(x) = g(f(x)) = g(2x + 5)$ $(g \circ f)(x) = (2x + 5)^2 - 3$ $(g \circ f)(x) = 4x^2 + 20x + 25 - 3 = 4x^2 + 20x + 22$ To evaluate at $x = 1$: $(g \circ f)(1) = 4(1)^2 + 20(1) + 22 = 4 + 20 + 22 = 46$

To find the composite function $g(f(x))$, substitute the entire expression for $f(x)$ into every $x$ in the function $g(x)$. Then simplify the resulting expression.

Find the inverse function $f^{-1}(x)$ for the function $f(x) = \frac{3x - 1}{x + 2}$, where $x \neq -2$.

1. Let $y = \frac{3x - 1}{x + 2}$
2. Swap $x$ and $y$: $x = \frac{3y - 1}{y + 2}$
3. Multiply by $(y + 2)$: $x(y + 2) = 3y - 1$ $xy + 2x = 3y - 1$
4. Rearrange to isolate $y$: $xy - 3y = -2x - 1$ $y(x - 3) = -(2x + 1)$ $y = \frac{-(2x + 1)}{x - 3} = \frac{2x + 1}{3 - x}$
5. Therefore, $f^{-1}(x) = \frac{2x + 1}{3 - x}$, $x \neq 3$.

To find the inverse algebraically, replace $f(x)$ with $y$, interchange $x$ and $y$, and then solve the resulting equation for $y$ in terms of $x$.

A function $h(x)$ has a domain $D = \{x \in \mathbb{R} : x \geq 0\}$ and range $R = \{y \in \mathbb{R} : y \geq 5\}$. State the domain and range of $h^{-1}(x)$.

Domain of $h^{-1} = \text{Range of } h = \{x \in \mathbb{R} : x \geq 5\}$ Range of $h^{-1} = \text{Domain of } h = \{y \in \mathbb{R} : y \geq 0\}$

Because the inverse function reflects the original function over $y=x$, the sets for the domain and range are swapped.

1. Find the composite expression: $(f \circ g)(x) = f(g(x)) = f(2x + 1) = e^{2x + 1}$
2. Set the expression equal to 5: $e^{2x + 1} = 5$
3. Take the natural logarithm of both sides: $\ln(e^{2x + 1}) = \ln(5)$ $2x + 1 = \ln(5)$
4. Solve for $x$: $2x = \ln(5) - 1$ $x = \frac{\ln(5) - 1}{2} \approx 0.305$

The composition involves substituting the linear function $g(x)$ into the exponent of the natural exponential function $f(x)$. The resulting equation is solved using logarithms.

1. Write the function as $y$: $y = \sqrt{x - 3}$
2. Swap $x$ and $y$: $x = \sqrt{y - 3}$
3. Solve for $y$: Square both sides: $x^2 = y - 3$ $y = x^2 + 3$ So, $f^{-1}(x) = x^2 + 3$.
4. Determine the domain: The range of $f(x) = \sqrt{x - 3}$ is $y \geq 0$. Therefore, the domain of $f^{-1}(x)$ is $x \geq 0$.

To find the inverse, we isolate the variable that was originally the input. Because the original function's output is always non-negative (square root), the inverse function is only defined for $x \geq 0$.