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# Mathematics I · Week 5 — Composition, inverse, and exponentials Deep study for Quiz 2 week 5. Master composition order first, then invertibility, then exponential growth rules.

Mathematics I · Week 5 — Composition, inverse, and exponentials

Deep study for Quiz 2 week 5. Master composition order first, then invertibility, then exponential growth rules.

Week map

Vertical/horizontal line tests → composition

(f \circ g)

→ inverse

f^{-1}

→ exponential

a^x

→ domain discipline throughout.

Composition notation

$(f \circ g)(x)$ → “ $f$ composed with $g$ at $x$ ” → apply $g$ first, then $f$ → $(f \circ g)(x) = f(g(x))$ .
$f(g(x))$ → inner function output becomes outer input → parentheses matter when substituting.
Domain of composition → all $x$ where $g(x)$ is in domain of $f$ .

Mini-example:

f(x) = 2x + 1

g(x) = x^2

. Then

(f \circ g)(3) = f(g(3)) = f(9) = 19

. But

(g \circ f)(3) = g(f(3)) = g(7) = 49

— order changes the answer.

Vertical and horizontal line tests

Vertical line test: graph is a function iff no vertical line hits twice.
Horizontal line test: function is injective (one-to-one) iff no horizontal line hits twice — needed for invertible function on its domain.

Inverse notation

$f^{-1}(y)$ → input that $f$ sends to $y$ → not the same as $1/f(x)$ .
Inverse swaps roles: if $f(a) = b$ then $f^{-1}(b) = a$ .
Domain of $f^{-1}$ = range of $f$ ; range of $f^{-1}$ = domain of $f$ .
Graph of $f^{-1}$ reflects graph of $f$ over line $y = x$ .

Mini-example:

f(x) = 3x - 2

\mathbb{R}

. Swap:

x = 3y - 2

, solve

y = \frac{x+2}{3}

. So

f^{-1}(x) = \frac{x+2}{3}

. Check:

f^{-1}(5) = 7/3

and

f(7/3) = 5

Exponential notation

$y = a^x$ → exponential with base $a$ → $a > 0$ , $a \neq 1$ .
$a^0 = 1$ for any valid base; $a^{m+n} = a^m \cdot a^n$ ; $(a^m)^n = a^{mn}$ .
$a^x > 0$ always — range is $(0, \infty)$ .

Mini-example:

2^3 = 8

;

2^{-1} = 1/2

;

(1/2)^2 = 1/4

. Base between 0 and 1 gives decay; base

> 1

gives growth.

Trap:

x^a

(power function) vs

a^x

(exponential) — variable position differs.

Pattern families

Easy — Evaluate composition

Substitute entire inner expression into outer. Simplify stepwise. Watch signs when replacing

x

inside

f

Medium — Find inverse of linear or simple rational

Swap

x

and

y

, solve for

y

. State domain of inverse explicitly. Verify

f(f^{-1}(x)) = x

on valid domain.

Hard — Invertibility and restricted domains

Use horizontal line test or monotonicity. Square root and squaring need domain restriction (

x \geq 0

) for bijection. Composition with inverse simplifies to identity only on matched domains.

Worked mini-examples

Example 1 — Composition order.

f(x) = x + 4

g(x) = 2x

(f \circ g)(5) = f(10) = 14

(g \circ f)(5) = g(9) = 18

Example 2 — Nested formula.

f(x) = x^2 + 1

g(x) = x - 3

(f \circ g)(x) = (x-3)^2 + 1 = x^2 - 6x + 10

Example 3 — Inverse of linear.

f(x) = \frac{x-1}{2}

. Swap:

x = \frac{y-1}{2}

→

2x = y - 1

→

f^{-1}(x) = 2x + 1

Example 4 — Exponential evaluation.

3^{2} = 9

;

3^{-2} = 1/9

;

3^{1/2} = \sqrt{3}

Example 5 — Not invertible without restriction.

f(x) = x^2

on all

\mathbb{R}

fails horizontal line test (

f(2) = f(-2) = 4

). Restrict to

x \geq 0

to define inverse

f^{-1}(x) = \sqrt{x}

Traps

Writing $(f \circ g)(x)$ as $f(x) \cdot g(x)$ .
Treating $f^{-1}$ as $1/f(x)$ .
Taking domain of $f^{-1}$ equal to domain of $f$ .
Applying outer function before inner in composition.
Confusing exponential base with exponent: $2^x$ vs $x^2$ .

Diagnostic (try yourself)

If $f(x) = 3x - 1$ and $g(x) = x^2$ , find $(g \circ f)(2)$ and $(f \circ g)(2)$ .
Find $f^{-1}(x)$ for $f(x) = 5x + 4$ . What is the domain of $f^{-1}$ if domain of $f$ is $\mathbb{R}$ ?
Is $f(x) = |x|$ injective on $\mathbb{R}$ ? What restriction makes it bijective from $\mathbb{R}$ to $[0,\infty)$ ?
Simplify $(f \circ g)(x)$ when $f(x) = \sqrt{x}$ and $g(x) = x + 5$ . State the domain of the composition.
Evaluate $2^{3}$ and $3^{2}$ . Which is larger? What is $2^{-3}$ ?