Functions - Functions, domain, range, graph

Given $f(x) = \frac{2x + 1}{x - 3}$, find the domain and range of $f$.

For the domain, the denominator cannot be zero: $x - 3 \neq 0 \Rightarrow x \neq 3$. Thus, the domain is $x \in \mathbb{R}, x \neq 3$. To find the range, we look for the horizontal asymptote. As $x \to \infty$, $f(x) \to \frac{2x}{x} = 2$. Thus, $y \neq 2$. The range is $y \in \mathbb{R}, y \neq 2$.

The domain is restricted by the vertical asymptote at $x=3$. The range is restricted by the horizontal asymptote at $y=2$.

Let $f(x) = e^{2x} - 5$. Find the expression for $f^{-1}(x)$ and state its domain.

1. Replace $f(x)$ with $y$: $y = e^{2x} - 5$.
2. Swap $x$ and $y$: $x = e^{2y} - 5$.
3. Solve for $y$: $x + 5 = e^{2y}$ $\ln(x + 5) = 2y$ $y = \frac{1}{2}\ln(x + 5)$. So, $f^{-1}(x) = \frac{1}{2}\ln(x + 5)$. The domain of $f^{-1}$ is the range of $f$. Since $e^{2x} > 0$, $f(x) > -5$. Thus, the domain of $f^{-1}$ is $x > -5$.

To find the inverse, we swap variables and solve for $y$. The domain of the logarithmic inverse is restricted to values that make the argument positive.

If $f(x) = x^2 + 3$ and $g(x) = \sqrt{x - 1}$, find $(f \circ g)(x)$ and its domain.

$(f \circ g)(x) = f(g(x)) = f(\sqrt{x - 1}) = (\sqrt{x - 1})^2 + 3 = x - 1 + 3 = x + 2$. For the domain, $g(x)$ must be defined, so $x - 1 \geq 0 \Rightarrow x \geq 1$. Even though the simplified expression $x+2$ looks like it accepts all $x$, the domain is restricted by the inner function $g(x)$. Domain: $x \geq 1$.

Composition requires the input to be valid for the inner function first.

1. For the domain, the expression inside the square root must be non-negative: $4 - x \geq 0 \Rightarrow x \leq 4$
2. For the range, since $\sqrt{4 - x} \geq 0$, then $-\sqrt{4 - x} \leq 0$. Adding 3 gives: $f(x) \leq 3$ Thus, Domain is $(-\infty, 4]$ and Range is $(-\infty, 3]$.

The graph starts at the endpoint $(4, 3)$ and moves to the left and downwards. The $x$-values are all numbers less than or equal to 4, and the $y$-values are all numbers less than or equal to 3.

1. Vertical Asymptote: Set denominator to zero: $x + 2 = 0 \Rightarrow x = -2$
2. Horizontal Asymptote: As $x \rightarrow \pm\infty$, $\frac{1}{x+2} \rightarrow 0$: $y = -1$
3. Domain: $x \in \mathbb{R}, x \neq -2$
4. Range: $y \in \mathbb{R}, y \neq -1$

The graph is a hyperbola translated 2 units left and 1 unit down. The domain and range exclude the values where the asymptotes are located.