Composite Functions
A shop in Cork takes €10 off a jacket and then 20% off the reduced price. Two steps, one after the other. Is there a single rule that gives the final price straight from the original price? There is, and it is called a composite function.
Functions in a chain
Think of a function as a machine. You put in an input, and one output comes out. A composite function connects two machines: the output of the first goes straight in as the input of the second.
We write $f(g(x))$ and read it as "$f$ of $g$ of $x$". The function closest to $x$ acts first. So in $f(g(x))$ you apply $g$ first, then $f$.
For a number, just work from the inside out.
Given $f(x) = 2x + 3$ and $g(x) = x^2$, find: a) $f(g(2))$ b) $g(f(2))$
Given $f(x) = 2x + 3$ and $g(x) = x^2$, find:
- a) $f(g(2))$
- b) $g(f(2))$
Given $h(x) = x^2 + 1$ and $k(x) = x + 4$, find $h(k(-1))$ and $k(h(-1))$.
Given $h(x) = x^2 + 1$ and $k(x) = x + 4$, find $h(k(-1))$ and $k(h(-1))$.
Finding the formula
To get a rule for the whole composite, replace every $x$ in the outer function by the whole expression of the inner function.
Composites suit real two-step situations. Back in the shop: $d(p) = p - 10$ takes off €10 and $s(p) = 0.8p$ takes off 20%. The final price is $s(d(p)) = 0.8(p - 10) = 0.8p - 8$. A jacket at €85 costs $0.8(85) - 8 = €60$. Back to the shop: one rule, $0.8p - 8$, does both discounts at once.
Given $f(x) = x - 5$ and $g(x) = 3x$, write down expressions for: a) $f(g(x))$ b) $g(f(x))$
Given $f(x) = x - 5$ and $g(x) = 3x$, write down expressions for:
- a) $f(g(x))$
- b) $g(f(x))$
Given $f(x) = 3x - 2$ and $g(x) = x^2 + 1$: a) Write $f(g(x))$ in the form $ax^2 + b$. b) Write $g(f(x))$ in the form $px^2 + qx + r$. c) Solve $f(g(x)) = 49$.
Given $f(x) = 3x - 2$ and $g(x) = x^2 + 1$:
- a) Write $f(g(x))$ in the form $ax^2 + b$.
- b) Write $g(f(x))$ in the form $px^2 + qx + r$.
- c) Solve $f(g(x)) = 49$.
A shop in Cork offers a discount of €10 off any item and then a further $20\%$ off the reduced price. Let $d(p) = p - 10$ and $s(p) = 0.8p$. a) Write the final price…
A shop in Cork offers a discount of €10 off any item and then a further $20\%$ off the reduced price. Let $d(p) = p - 10$ and $s(p) = 0.8p$.
- a) Write the final price after both discounts as a composite function of $p$ in the form $s(d(p))$ and simplify it.
- b) Find the final price of a jacket originally priced at €85.
- c) Find the original price of a coat whose final price is €52.
Equations with composites
Once you have the formula, an equation such as $f(g(x)) = 49$ is just an ordinary equation in $x$. Simplify the composite first, then solve.
When a quadratic composite is set equal to a number, you may get two solutions, so check for both. If the equation includes an unknown constant, such as $f(x) = 2x + a$, substitute and compare terms: if $f(g(x)) = g(f(x))$ for all $x$, the coefficients of $x$ and the constant terms must match.
Given $f(x) = x + 2$ and $g(x) = x^2 - 3$, solve the equation $g(f(x)) = f(g(x))$.
Given $f(x) = x + 2$ and $g(x) = x^2 - 3$, solve the equation $g(f(x)) = f(g(x))$.
Given $f(x) = 2x + a$ and $g(x) = 3x - 1$, where $a$ is a constant, it is known that $f(g(x)) = g(f(x))$ for all $x$. a) Find the value of $a$. b) Verify your answer by…
Given $f(x) = 2x + a$ and $g(x) = 3x - 1$, where $a$ is a constant, it is known that $f(g(x)) = g(f(x))$ for all $x$.
- a) Find the value of $a$.
- b) Verify your answer by calculating both composites at $x = 3$.
Functions that undo each other
Let $f(x) = \dfrac{x+1}{2}$ and $g(x) = 2x - 1$. Then $f(g(x)) = \dfrac{(2x - 1) + 1}{2} = \dfrac{2x}{2} = x$, and $g(f(x)) = 2\left(\dfrac{x+1}{2}\right) - 1 = x + 1 - 1 = x$.
Both composites give back exactly $x$. Whatever $f$ does, $g$ reverses, and the other way round. We say $f$ and $g$ are inverse functions: each one undoes the other.
You can also compose a function with itself. For $f(x) = \dfrac{x+1}{2}$:
$f(f(x)) = \dfrac{\frac{x+1}{2} + 1}{2} = \dfrac{x + 3}{4}$.
To solve $f(f(x)) = 5$: $\dfrac{x+3}{4} = 5$, so $x + 3 = 20$ and $x = 17$.
Given $f(x) = \dfrac{x+1}{2}$ and $g(x) = 2x - 1$: a) Show that $f(g(x)) = x$ and $g(f(x)) = x$. b) Explain what this tells you about the relationship between $f$ and…
Given $f(x) = \dfrac{x+1}{2}$ and $g(x) = 2x - 1$:
- a) Show that $f(g(x)) = x$ and $g(f(x)) = x$.
- b) Explain what this tells you about the relationship between $f$ and $g$.
- c) Find $f(f(x))$ and solve $f(f(x)) = 5$.
Given $f(x) = x^2 - 4x$ and $g(x) = x + 2$: a) Show that $f(g(x)) = x^2 - 4$. b) Find the values of $x$ for which $f(g(x)) = g(f(x))$. c) Find the minimum value of…
Given $f(x) = x^2 - 4x$ and $g(x) = x + 2$:
- a) Show that $f(g(x)) = x^2 - 4$.
- b) Find the values of $x$ for which $f(g(x)) = g(f(x))$.
- c) Find the minimum value of $g(f(x))$ and the value of $x$ at which it occurs.