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Week 5: Graph Tests, Composition, Inverses & Exponentials

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Week 5: Graph Tests, Composition, Inverses & Exponentials

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15 min read

2026-09-12

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Mathematics I — Week 5: Graph Tests, Composition, Inverses & Exponentials

IITM’s Week 5 covers horizontal and vertical line tests, exponential functions, composite functions, and inverse functions.

TL;DR

Week 5 deepens the idea of a function:

\text{function} \rightarrow \text{graph test} \rightarrow \text{composition} \rightarrow \text{one-one} \rightarrow \text{inverse} \rightarrow \text{exponential functions}

The highest-value ideas are:

vertical line test → is it a function?
horizontal line test → is it one-one?
composition: $(f\circ g)(x)=f(g(x))$
inverse: $f^{-1}(x)$
invertibility requires one-one behavior
exponential functions: $a^x$

1. Function recap

A function:

f:A\to B

assigns every input in

A

exactly one output in

B

The three main sets are:

domain
codomain
range

Always:

\text{Range}\subseteq \text{Codomain}.

2. Function from a graph

Suppose a relation is drawn in the

xy

-plane.

How do we know whether it represents:

y=f(x)?

Use the vertical line test.

3. Vertical line test

A graph represents a function of

x

if every vertical line intersects it at most once.

Why?

A vertical line corresponds to fixing one value of

x

If the graph intersects twice, the same input

x

has two different outputs.

That violates the definition of a function.

4. Example: parabola

Consider:

y=x^2.

For each

x

, there is exactly one

y

Every vertical line intersects once.

So:

y=x^2

is a function.

5. Example: circle

Consider:

x^2+y^2=1.

At:

x=0,

we get:

y=\pm1.

So one input:

x=0

has two outputs.

Therefore the circle fails the vertical line test.

It is not a function:

y=f(x)

over the whole circle.

6. Horizontal line test

The horizontal line test asks whether a function is one-one / injective.

A function is one-one if every horizontal line intersects its graph at most once.

Why?

A horizontal line fixes one output

y

If it intersects twice, then two different inputs produce the same output.

That means:

f(x_1)=f(x_2)

with:

x_1\neq x_2.

So the function is not injective.

7. Example

For:

f(x)=x^2,

we have:

f(2)=4

and:

f(-2)=4.

Thus it is not one-one on:

\mathbb R.

The graph fails the horizontal line test.

8. Restricting the domain

A function may fail to be one-one on a large domain but become one-one after restricting the domain.

For:

f(x)=x^2,

if we restrict to:

x\geq0,

then the function becomes one-one.

Similarly on:

x\leq0.

This matters enormously for inverse functions.

9. Composition of functions

Suppose:

f:A\to B

and:

g:B\to C.

Then the composition:

g\circ f

maps:

A\to C.

Definition:

\boxed{ (g\circ f)(x)=g(f(x)) }

Read:

$g$ composed with $f$ .

Important:

Apply $f$ first, then $g$ .

10. Order matters

In:

(g\circ f)(x),

the inner function acts first.

So:

x \overset{f}{\longrightarrow} f(x) \overset{g}{\longrightarrow} g(f(x)).

This is similar to nested function calls in programming.

11. Example

Let:

f(x)=2x+1

and:

g(x)=x^2.

Then:

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

Since:

f(x)=2x+1,

we substitute into

g

g(2x+1)=(2x+1)^2.

Thus:

\boxed{ (g\circ f)(x)=(2x+1)^2 }

12. Reverse composition

Now compute:

(f\circ g)(x).

That means:

f(g(x)).

Since:

g(x)=x^2,

we get:

f(x^2)=2x^2+1.

Therefore:

\boxed{ (f\circ g)(x)=2x^2+1 }

These are different:

g\circ f\neq f\circ g

in general.

Composition is not commutative.

13. Domain of a composition

This is a subtle but important point.

For:

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

x

must:

belong to the domain of $g$
produce a value $g(x)$ that belongs to the domain of $f$

So the composition domain can be smaller than either function's apparent domain.

14. Example: domain restriction in composition

Let:

g(x)=x-3

and:

f(x)=\sqrt{x}.

Then:

(f\circ g)(x) = \sqrt{x-3}.

We require:

x-3\geq0.

Thus:

\boxed{x\geq3}

is the domain.

15. Identity function

The identity function is:

\boxed{ I(x)=x }

It changes nothing.

For a function

f

f\circ I=f

and:

I\circ f=f.

The identity function becomes central when defining inverses.

16. What is an inverse function?

An inverse function reverses the action of the original function.

If:

f(a)=b,

then:

f^{-1}(b)=a.

So:

f:A\to B

is reversed by:

f^{-1}:B\to A.

17. Inverse notation warning

f^{-1}(x)

does not mean:

\frac1{f(x)}.

These are entirely different.

For example, if:

f(x)=2x,

then:

f^{-1}(x)=\frac x2.

But:

\frac1{f(x)}=\frac1{2x}.

Do not confuse them.

18. Inverse as composition

f^{-1}

truly is the inverse of

f

, then:

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

and:

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

So:

f^{-1}\circ f=I

and:

f\circ f^{-1}=I.

19. When does an inverse exist?

For a genuine inverse:

f^{-1}:B\to A,

the function must be bijective:

injective
surjective

Why?

Each output must correspond to exactly one input.

If two inputs share one output, reversing the mapping would give one input with two outputs.

Not allowed.

20. Graph criterion for inverses

A function has an inverse on its domain/range if it is one-one.

Graphically:

It must pass the horizontal line test.

So the horizontal line test is really an inverse-existence test.

21. Finding an inverse algebraically

Suppose:

f(x)=2x+3.

Step 1

Write:

y=2x+3.

Step 2

Swap

x

and

y

x=2y+3.

Step 3

Solve for

y

x-3=2y

y=\frac{x-3}{2}.

Therefore:

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

22. Why swapping works

Original function maps:

x\to y.

Inverse maps:

y\to x.

So swapping

x

and

y

literally reverses input and output roles.

23. Verify the inverse

For:

f(x)=2x+3

and:

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

compute:

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

Then:

f\left(\frac{x-3}{2}\right) = 2\left(\frac{x-3}{2}\right)+3.

Thus:

=x-3+3

=x.

Correct.

24. Inverse graph geometry

The graphs of:

y=f(x)

and:

y=f^{-1}(x)

are reflections of each other across the line:

\boxed{y=x}

Why?

Inverse swaps:

(x,y) \leftrightarrow (y,x).

Reflection across

y=x

does exactly that.

25. Domain/range swap under inverse

If:

f:A\to B,

then:

f^{-1}:B\to A.

Therefore:

\boxed{ \text{Domain of }f^{-1} = \text{Range of }f }

and:

\boxed{ \text{Range of }f^{-1} = \text{Domain of }f }

26. Quadratic inverse problem

Consider:

f(x)=x^2.

On:

\mathbb R,

it is not one-one.

So it does not have an inverse function on all real numbers.

Why?

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

27. Restricting the domain

Restrict:

f(x)=x^2,\qquad x\geq0.

Now it is one-one.

Then:

y=x^2

Swap:

x=y^2.

Since original domain required nonnegative input, inverse output must be nonnegative:

y=\sqrt{x}.

Thus:

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

28. Why not $\pm\sqrt{x}$ ?

A function must give exactly one output per input.

So:

\pm\sqrt{x}

would not define a function.

The domain restriction determines which branch is selected.

This is a very common trap.

29. Exponential functions

An exponential function has the form:

\boxed{ f(x)=a^x }

where typically:

a>0,\qquad a\neq1.

The variable appears in the exponent.

This distinguishes exponentials from polynomials.

30. Polynomial vs exponential

Polynomial:

x^3

Variable in the base.

Exponential:

3^x

Variable in the exponent.

These behave very differently.

31. Basic exponential properties

For:

a>0,

we have:

a^0=1.

So every basic exponential graph passes through:

\boxed{(0,1)}

Also:

a^1=a.

So it passes through:

(1,a).

32. Exponent laws

For suitable

a

a^m a^n=a^{m+n}

\frac{a^m}{a^n}=a^{m-n}

(a^m)^n=a^{mn}

a^{-n}=\frac1{a^n}

a^{1/n}=\sqrt[n]{a}.

These are core algebra tools.

33. Exponential growth

If:

a>1,

then:

f(x)=a^x

is increasing.

Example:

2^x.

x

increases:

2^x

grows rapidly.

34. Exponential decay

If:

0<a<1,

then:

a^x

is decreasing.

Example:

\left(\frac12\right)^x.

x

increases, the output approaches zero.

35. Domain of exponential function

For:

f(x)=a^x,\qquad a>0,

the domain is:

\boxed{\mathbb R}

because real powers are defined appropriately for positive base

a

36. Range

For:

a^x,

outputs are always positive:

a^x>0.

Therefore:

\boxed{ \text{Range}=(0,\infty) }

Never:

0

and never negative.

37. Horizontal asymptote

As exponential decay approaches zero, it never actually reaches zero.

The line:

y=0

acts as a horizontal asymptote.

Similarly, for growth:

a^x\to0

as:

x\to-\infty.

38. Exponential graph summary

For:

a>1:

passes through $(0,1)$
always positive
increasing
approaches $0$ to the left
grows rapidly to the right

For:

0<a<1:

passes through $(0,1)$
always positive
decreasing
grows large to the left
approaches $0$ to the right

39. Exponential function is one-one

For:

a>0,\quad a\neq1,

the exponential function is monotonic.

Therefore it passes the horizontal line test.

So it has an inverse.

That inverse is the logarithm, which becomes Week 6.

This is a key conceptual bridge.

40. Solving simple exponential equations

Suppose:

2^x=8.

Since:

8=2^3,

we get:

x=3.

Similarly:

3^{2x}=27.

Since:

27=3^3,

we get:

2x=3

so:

x=\frac32.

41. Same-base principle

If:

a^u=a^v

with:

a>0,\quad a\neq1,

then:

\boxed{u=v}.

This works because the exponential function is one-one.

So injectivity from Week 1 directly justifies this algebra rule.

42. Example

Solve:

5^{2x-1}=5^{x+4}.

Since same valid base:

2x-1=x+4.

Thus:

x=5.

43. Exponential transformations

Consider:

f(x)=a^x.

Then:

a^{x-h}

shifts the graph horizontally by

h

And:

a^x+k

shifts vertically by

k

So:

f(x)=2^{x-3}+5

has the base graph shifted:

right $3$
up $5$

Its horizontal asymptote becomes:

y=5.

44. Reflection

-a^x

reflects the graph across the

x

-axis.

Meanwhile:

a^{-x}

reflects across the

y

-axis.

And:

a^{-x} = \left(\frac1a\right)^x.

This is why exponential growth and decay graphs are reflections.

45. Example

2^{-x} = \left(\frac12\right)^x.

Thus the decay function:

\left(\frac12\right)^x

is just the reflection of:

2^x

across the

y

-axis.

46. Composition and inverse interaction

Suppose:

f(x)=2x+3.

Its inverse is:

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

Then:

f^{-1}\circ f=I

because:

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

So inverse functions are literally functions that undo composition.

47. Inverse of a composition

If suitable inverses exist:

\boxed{ (f\circ g)^{-1} = g^{-1}\circ f^{-1} }

Notice the order reverses.

Why?

If:

g

happens first and then:

f,

to undo the process:

undo $f$ first
then undo $g$

This is exactly like reversing a sequence of operations.

48. Example intuition

Suppose a process:

multiply by $2$
add $3$

This gives:

f(x)=2x+3.

To undo:

subtract $3$
divide by $2$

Reverse order.

That is the essence of inverse functions.

49. One-one vs onto vs inverse

This distinction is worth keeping clean.

Injective

Different inputs → different outputs.

Surjective

Every codomain value is reached.

Bijective

Both.

Invertible

For a full inverse:

B\to A,

you need bijectivity.

If discussing inverse on the actual range, injectivity is the key condition.

Common trap: vertical vs horizontal line tests

Vertical line test

Asks:

Is this relation a function?

Horizontal line test

Asks:

Is this function one-one?

Do not swap them.

Common trap: composition order

(f\circ g)(x)

means:

f(g(x)),

not:

g(f(x)).

The function closest to

x

acts first.

Common trap: inverse notation

Again:

f^{-1}(x) \neq \frac1{f(x)}.

For:

f(x)=x+1,

inverse:

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

Reciprocal:

\frac1{x+1}.

Completely different objects.

Common trap: inverse domain

Suppose:

f(x)=x^2,\quad x\geq0.

Then:

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

But the inverse domain is:

x\geq0.

You cannot simply ignore domain/range restrictions.

Common trap: base of exponential

For a standard real exponential:

a^x,

require:

a>0

and:

a\neq1.

If:

a=1,

then:

1^x=1,

which is just a constant function.

55. Why exponential growth is special

Compare:

x^{10}

with:

2^x.

For sufficiently large

x

2^x

eventually grows faster than:

x^{10}.

More generally, exponential functions eventually dominate fixed-degree polynomials.

This becomes important in algorithms, statistics, growth models, and later mathematics.

56. Function transformations recap

For a base function:

y=f(x),

Horizontal shift

f(x-h)

→ right by

h

Vertical shift

f(x)+k

→ up by

k

Vertical reflection

-f(x)

→ reflect across

x

-axis.

Horizontal reflection

f(-x)

→ reflect across

y

-axis.

Vertical scaling

af(x)

→ stretch/compress vertically.

This transformation language applies far beyond exponentials.

57. Full example: analyze a transformed exponential

Consider:

f(x)=3^{x-2}+4.

Start from:

y=3^x.

Transformations:

x-2

means shift right by

2

Then:

+4

means shift up by

4

Domain:

\mathbb R.

Original range:

(0,\infty).

After shifting up:

\boxed{(4,\infty)}.

Horizontal asymptote:

\boxed{y=4}.

58. Full inverse example

Let:

f(x)=3x-7.

Find inverse.

Start:

y=3x-7.

Swap:

x=3y-7.

Solve:

x+7=3y

y=\frac{x+7}{3}.

Thus:

\boxed{ f^{-1}(x)=\frac{x+7}{3} }

Verify:

f\left(\frac{x+7}{3}\right) = 3\left(\frac{x+7}{3}\right)-7 =x.

59. Full composition example

Let:

f(x)=x^2+1

and:

g(x)=2x-3.

Then:

(f\circ g)(x) = f(2x-3).

So:

=(2x-3)^2+1.

Expand:

4x^2-12x+9+1

thus:

\boxed{ 4x^2-12x+10 }

Reverse:

(g\circ f)(x) = g(x^2+1)

=2(x^2+1)-3

\boxed{ 2x^2-1 }

Clearly:

f\circ g\neq g\circ f.

60. Conceptual connection to programming

Function composition is closely related to nested function calls:


python
f(g(x))

means:

evaluate g(x)
pass result into f

Mathematical composition:

(f\circ g)(x)

is exactly the same conceptual pipeline.

That connection will be useful in Python too.

61. Conceptual connection to data science

A lot of data processing is composition:

\text{raw data} \to \text{clean} \to \text{transform} \to \text{normalize} \to \text{model}

Each step can be thought of as a function.

The full pipeline is a composition of functions.

So this is not abstract notation for its own sake.

62. Week 5 formula sheet

Composition

\boxed{ (f\circ g)(x)=f(g(x)) }

Identity

\boxed{ I(x)=x }

Inverse condition

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

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

Exponential

\boxed{ f(x)=a^x,\qquad a>0,\ a\neq1 }

Exponential domain

\boxed{\mathbb R}

Exponential range

\boxed{(0,\infty)}

Same-base principle

\boxed{ a^u=a^v \Rightarrow u=v }

for valid

a\neq1

63. What you should actually master

By the end of Week 5, you should be able to:

Apply the vertical line test.
Apply the horizontal line test.
Determine whether a function is one-one.
Understand why domain restriction may be necessary.
Compute: $f(g(x)).$
Distinguish: $f\circ g$ from: $g\circ f.$
Determine domains of compositions.
Find inverses algebraically.
Verify inverses using composition.
Swap domain and range correctly.
Understand inverse graphs as reflections across:

y=x.

Recognize exponential functions.
Distinguish exponential growth from decay.
Use exponent laws.
Solve simple same-base exponential equations.
Analyze basic exponential graph transformations.

If those are comfortable, Week 5 is under control.

64. Concept map


text
Functions
│
├── Graph
│   ├── vertical line test
│   └── horizontal line test
│
├── Composition
│   └── f(g(x))
│
├── One-one
│   └── invertibility
│
├── Inverse
│   ├── swap x/y
│   ├── domain ↔ range
│   └── reflection across y=x
│
└── Exponential
    ├── growth
    ├── decay
    └── inverse → logarithm

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:

Question :: Answer.

Vertical Line Test determines :: Whether a curve in the plane represents a valid function
Horizontal Line Test determines :: Whether a function is injective (and therefore invertible)
Composite function $(f \circ g)(x)$ definition :: $f(g(x))$
Graph of $f^{-1}(x)$ relative to $f(x)$ :: Reflected across the line $y = x$
Condition for existence of inverse function $f^{-1}$ :: The function must be bijective (one-to-one and onto)
Exponential function $f(x) = a^x$ domain and range ( $a > 0, a \neq 1$ ) :: Domain: $\mathbb{R}$ , Range: $(0, \infty)$

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.

Question 1

Let

f(x) = 2x + 1

and

g(x) = x^2 - 3

. What is

(f \circ g)(2)

NAT

3 Unit Assessment

Question 2

Which test is used on the graph of a function to determine whether it has an inverse function?

MCQ

3 Unit Assessment

A. Vertical Line Test

B. Horizontal Line Test

C. Integral Test

D. Ratio Test