Topic 2.7 Notes – Composition of Functions

Nesting one function inside another

If you see

$$(f\circ g)(x)=f(g(x)),$$

read it as “$g$ happens first, then $f$”.

That means the path is

$$x \to g(x) \to f(g(x)).$$

• Inside function means the one closest to $x$. Here, that is $g$.
• Outside function means the one applied after that. Here, that is $f$.
• $f(g(x))$ is function notation. It does not mean $f(x)g(x)$.

Composition is useful when two quantities connect through a middle quantity.
Example: time $\to$ radius $\to$ area.

• If $r(t)=1+2t$ and $A(r)=\pi r^2$, then

$$(A\circ r)(t)=A(r(t))=\pi(1+2t)^2.$$

The domain of $f\circ g$ needs both steps to work:

• $x$ must be in the domain of $g$
• $g(x)$ must be in the domain of $f$

Also, order matters. Usually,

$$f\circ g \ne g\circ f.$$

One special function does nothing to the input. The identity function is

$$I(x)=x.$$

So

$$I\circ g=g\circ I=g.$$

Evaluating and Constructing Composites

Evaluating values

For a number input, do the inside first.

If $f(x)=x^2+1$ and $g(x)=3x-2$, then

$$(f\circ g)(2)=f(g(2)).$$

Compute:

1. $g(2)=3(2)-2=4$
2. $f(4)=4^2+1=17$

So $(f\circ g)(2)=17$.

If there are three functions, work from the innermost outward.

Constructing a formula

To build a rule for $f\circ g$, replace every $x$ in $f(x)$ with $g(x)$.

$$(f\circ g)(x)=f(g(x))$$

Example: $f(x)=x^2+5$, $g(x)=2x-1$

$$(f\circ g)(x)=(2x-1)^2+5=4x^2-4x+6.$$

The graph helps you see that the composite uses the output of $g$ as the input to $f$.

Parentheses matter a lot, especially with powers and negatives.

Graphs of $f$, $g$, and $f\circ g$

From tables, graphs, and situations

• Tables: find $g(x)$ in the first table, then use that output as the input in the $f$ table.
• If that input is missing in the second table, the composite may be undefined or cannot be determined.
• Graphs: read $g(a)$, then go to the graph of $f$ and evaluate at that value.
• Watch for holes, endpoints, and restricted domains on both graphs.
• Verbal situations: let units guide the order. If miles $\to$ gallons $\to$ cost, that tells you which function is inside.

Domain and Order Matter

This is where students lose easy points.

The full rule is

• an input survives only if $g$ accepts it and
• the result $g(x)$ is allowed in $f$

Common restrictions:

• denominator cannot be zero
• even-root radicand must be nonnegative
• context may restrict values too

A simplified formula can hide excluded values.

Example:

$$f(x)=\frac{x^2-1}{x-1},\quad g(x)=x$$ Then $$(f\circ g)(x)=\frac{x^2-1}{x-1}=x+1,$$ but $x=1$ is still excluded.

Reversing order can change formula, domain, range, graph, and even units.

Decomposing a Function into Simpler Functions

Decomposition means reversing composition.

Suppose

$$h(x)=\sqrt{3x^2+7}.$$

A clean decomposition is

• $g(x)=3x^2+7$
• $f(x)=\sqrt{x}$

Then

$$h(x)=f(g(x)).$$

Look for the natural inside piece, often a repeated expression or the part “done first.”

Sometimes you can use more than two layers. Decompositions are often not unique. What matters is that recomposing gives the original function back, with the same domain.

Transformations as Compositions

This is the connection to graph transformations.

$$f(x)+k=(A_k\circ f)(x), \quad A_k(x)=x+k$$ This changes outputs, so it is a vertical translation.

$$f(x+k)=(f\circ A_k)(x)$$ This changes inputs, so it is a horizontal translation.

$$kf(x)=(M_k\circ f)(x), \quad M_k(x)=kx$$ This is a vertical dilation. If $k<0$, it also reflects across the $x$-axis.

$$f(kx)=(f\circ M_k)(x)$$ This is a horizontal dilation by factor $1/|k|$. If $k<0$, it also reflects across the $y$-axis.

The lock-in idea is simple:

• outside composition changes outputs
• inside composition changes inputs