unit 3.4 & 3.5- august 21st
Composition of Functions
- Core idea: when you see a composition like $f\circ g$ (i.e., $f(g(x))$), compute inside first and then apply the outer function.
- General rule for domains:
- For $f\circ g$, the domain is
\mathrm{Dom}(f\circ g)={x\in \mathrm{Dom}(g)\;|
\;g(x)\in \mathrm{Dom}(f)}. - Intuition: you must first ensure the inside function $g(x)$ is defined, and then ensure the value $g(x)$ lies in the domain of $f$.
- For $f\circ g$, the domain is
- Two common approaches to the domain of a composition:
- Approach A: find the domain of $g$ first, then impose the condition $g(x)\in \mathrm{Dom}(f)$.
- Approach B: analyze the combined expression and identify where any denominator becomes zero or any square root/domain restrictions fail, then convert to interval notation.
- Important note on inside-out reasoning: always start with the inside expression and work outward; this is the key to correctly evaluating nested functions.
Worked Examples
- Example 1: nested polynomials
- Given: $f(x) = x^2 - x$, $h(x) = 3x+2$ (so we want $f(h(1))$).
- Compute inside value: $h(1) = 3(1) + 2 = 5$.
- Compute outer value: $f(5) = 5^2 - 5 = 25 - 5 = 20$.
- Conclusion: $f(h(1)) = 20$.
- Takeaway: always evaluate the inside first, then substitute into the outer function.
- Example 2: different outer/inner functions
- Given: $f(x) = x + 5$, $h(x) = x^2 - 1$. Find $f(h(2))$.
- Compute inside value: $h(2) = 2^2 - 1 = 3$.
- Compute outer value: $f(3) = 3 + 5 = 8$.
- Conclusion: $f(h(2)) = 8$.
- Example 3: domain of a composition with rational functions
- Let $f(x) = \dfrac{5}{x-1}$ and $g(x) = \dfrac{4}{3x-2}$.
- Domain considerations:
- Domain of $g$: $3x-2 \neq 0 \Rightarrow x \neq \dfrac{2}{3}$.
- Domain of $f$: $x-1 \neq 0 \Rightarrow x \neq 1$.
- For $f\circ g$ to exist, we also need $g(x) \neq 1$ (to avoid division by zero in $f$):
\dfrac{4}{3x-2} \neq 1 \Rightarrow 4 \neq 3x-2 \Rightarrow x \neq 2. - Therefore, the domain of $f\circ g$ is
(-\infty, \tfrac{2}{3}) \cup (\tfrac{2}{3}, 2) \cup (2, \infty). - This illustrates the step of ensuring both the inner function is defined and the inner value avoids the outer function’s undefined points.
- Example 4: another domain example with square roots
- Let $f(x) = \sqrt{x+2}$ and $g(x) = \sqrt{3-x}$.
- Domain of $g$: need $3 - x \ge 0 \Rightarrow x \le 3$.
- For the composition $f\circ g$, require $g(x) \ge -2$ (the inside of $f$ must be nonnegative). Since $g(x) = \sqrt{3-x} \ge 0$, this is automatically satisfied.
- Therefore, the domain of $f\circ g$ is $(-\infty, 3]$.
- Example 5: decomposing a composition (two-function decomposition)
- Goal: express a function as a composition of two simpler functions.
- Given: $f(x) = \sqrt{5 - x^2}$.
- One valid decomposition: let $h(x) = 5 - x^2$ and $f{out}(u) = \sqrt{u}$, so that f(x) = f{out}(h(x)) = \sqrt{5 - x^2}.
- There can be multiple valid decompositions; any pair $(h, f_{out})$ that yields the original when composed is acceptable. The key idea is to separate the inside and the outer operation.
Notes on domains and graphs
- When a problem involves graphs of two functions, determine which is the inner function and which is the outer function.
- If a graph shows $f$ and $g$, identify which is applied first to avoid incorrect inside-out evaluation.
Decomposition and Transformation of Functions
Decomposing a given composition
- You are given a composite like $f\circ g$ and asked to write it as two functions.
- Two common strategies:
- Let the inner part be the inside of the composite, and assign the outer operation to the outer function.
- Example: If $f(x) = \sqrt{x}$ and the composition is $f(g(x)) = \sqrt{5 - x^2}$, you can set
- $h(x) = 5 - x^2$ and $g(x) = \sqrt{x}$ with a suitable outer function, or simply set the outer $f$ to be the square root and the inner to be $5 - x^2$ so that $f(h(x)) = \sqrt{5 - x^2}$.
- There are multiple valid decompositions; the idea is to identify an inner function and an outer function whose composition gives the original expression.
Transformations of graphs (basic moves)
- Vertical shifts (up/down): If you replace $f(x)$ by $f(x) + c$, the entire graph shifts up by $c$ units. If you replace by $f(x) - c$, it shifts down by $c$.
- Example: If $g(x) = f(x) - 3$, the graph shifts down by 3.
- Horizontal shifts (left/right): Inside the function, the sign is opposite to the shift.
- $f(x + a)$ shifts left by $a$.
- $f(x - a)$ shifts right by $a$.
- Example: $f(x - 3)$ shifts right by 3.
- Reflections
- Vertical reflection: $-f(x)$ reflects across the x-axis.
- Horizontal reflection: $f(-x)$ reflects across the y-axis.
- If a negative sign is outside the entire function, it’s a vertical reflection; a negative sign inside (on the x) is a horizontal reflection.
- Even and odd functions (parity)
- Even: $f(-x) = f(x)$ for all $x$ in the domain (symmetry about the y-axis).
- Odd: $f(-x) = -f(x)$ for all $x$ in the domain (symmetry about the origin).
- Example: $f(x) = x^2$ is even; $f(x) = x^3$ is odd.
- A function like $f(x) = x^3 + 2x$ is neither even nor odd, since $f(-x) = -x^3 - 2x
eq f(x)$ and $f(-x)
eq -f(x)$.
- Vertical stretch/compression
- If you multiply the function by a constant $a$ (i.e., consider $a f(x)$):
- If $|a| > 1$, the graph is stretched vertically by factor $|a|$.
- If $0 < |a| < 1$, the graph is compressed vertically by factor $|a|$.
- If $a < 0$, there is also a reflection about the x-axis in addition to the stretch/compression by $|a|$.
- Example: $g(x) = \tfrac{1}{2} f(x)$ is a vertical compression by a factor of $2$.
Quick practice reminders
- When shifting vertically, focus on how the y-values change (x-values stay the same).
- When shifting horizontally, focus on how the x-values change (y-values adapt to the new x).
- To determine even/odd, test with $f(-x)$ and compare to $f(x)$ and to $-f(x)$.
- For domain questions with square roots, ensure the inside of every square root is nonnegative; for rational functions, ensure denominators are nonzero; for nested compositions, ensure the inner function’s range lies in the outer function’s domain.
Summary of key formulas (LaTeX)
Domain of composition:
\mathrm{Dom}(f\circ g)={x\in \mathrm{Dom}(g)\mid g(x)\in \mathrm{Dom}(f)}.Horizontal shifts: f(x+a) \text{ shifts left by } a, \quad f(x-a) \text{ shifts right by } a.
Vertical shifts: f(x) + c \text{ shifts up by } c, \quad f(x) - c \text{ shifts down by } c.
Reflections:
- Vertical: -f(x) \text{ reflects across the } x ext{-axis}.
- Horizontal: f(-x) \text{ reflects across the } y ext{-axis}.
Parity:
- Even: $f(-x)=f(x)$; Odd: $f(-x)=-f(x)$.
Vertical stretch/compression:
- Multiply by $a$: y=a f(x);\quad |a|>1:\text{ stretch},\ 0<|a|<1:\text{ compression},\ a<0:\text{ reflection about x-axis in addition to stretch/compression.}
Example domain result (summary):
- If $g(x)=\dfrac{4}{3x-2}$ and $f(x)=\dfrac{5}{x-1}$, then
- $\mathrm{Dom}(g)={x\neq 2/3}$ and $\mathrm{Dom}(f)={x\neq 1}$.
- To have $f(g(x))$ defined, require $g(x)\neq 1$, which yields $x\neq 2$ as well as $x\neq 2/3$; hence
\mathrm{Dom}(f\circ g)=(-\infty,\tfrac{2}{3})\cup(\tfrac{2}{3},2)\cup(2,\infty).
Example domain result with square roots (summary):
- If $f(x)=\sqrt{x+2}$ and $g(x)=\sqrt{3-x}$, then
- Domain of $g$: $x\le 3$.
- For $f(g(x))=\sqrt{g(x)+2}$, require $g(x)\ge -2$; since $g(x)\ge 0$, this is always satisfied.
- Domain of $f\circ g$: $(-\infty, 3]$.