Functions Part 2 — Mathslive.ie

Section 1 of 12

Graphing a quadratic

Draw $f(x) = x^{2} - 2x - 3$ on the domain $-2 \le x \le 4$.

Build the table by working out $f(x)$ at each value of $x$:

$f(-2) = (-2)^{2} - 2(-2) - 3 = 4 + 4 - 3 = 5$

$f(-1) = 1 + 2 - 3 = 0$

$f(0) = -3 = -3$

$f(1) = 1 - 2 - 3 = -4$

$f(2) = 4 - 4 - 3 = -3$

$f(3) = 9 - 6 - 3 = 0$

$f(4) = 16 - 8 - 3 = 5$

Couples to plot:

$(-2,5),\; (-1,0),\; (0,-3),\; (1,-4),\; (2,-3),\; (3,0),\; (4,5)$

$f(x) = x^{2} - 2x - 3$ — the curve through all 7 couples

Notice the symmetry — the graph mirrors around $x = 1$. The lowest point is the minimum.

Section 2 of 12

Reading the graph

The same graph $f(x) = x^{2} - 2x - 3$ can answer lots of questions. Each one pins down either $x$ or $y$.

Two cues to remember

$f(x) = 0$ means $y = 0$, i.e. the $x$-axis. These are the roots.

$f(\text{number})$ means $x = \text{that number}$ — go up/down to the curve, read off $y$.

i Solve $x^{2} - 2x - 3 = 0$ from the graph

$f(x) = 0$ ⇒ look at the $x$-axis. The curve crosses at $x = -1$ and $x = 3$.

Roots — where the curve meets $y = 0$

Check algebraically by factorising:

$x^{2} - 2x - 3 = 0$

$(x + 1)(x - 3) = 0$

$x = -1,\; x = 3$ ✓

ii Find $f(-0.5)$ from the graph

$f(-0.5)$ ⇒ $x = -0.5$. Go up to the curve, then across to the $y$-axis.

Up from $x = -0.5$, then across — reads $y \approx -1.75$

Check by substituting:

$f(-0.5) = (-0.5)^{2} - 2(-0.5) - 3 = 0.25 + 1 - 3$

$= -1.75$ ✓

iii Find $x$ when $f(x) = 3$

$f(x) = 3$ ⇒ $y = 3$. Go across at $y = 3$, then down to the curve, then down to the $x$-axis.

Across at $y=3$ → down to curve → down to $x$-axis

$x \approx -1.65$ or $x \approx 3.65$

Section 3 of 12

Minimum point & axis of symmetry

To find the minimum of $f(x) = x^{2} - 2x - 3$ algebraically, complete the square.

$y = x^{2} - 2x - 3$

$y = (x^{2} - 2x + 1) - 3 - 1$ (add and subtract $1$)

$y = (x - 1)^{2} - 4$

Read off the minimum

If $y = (x - h)^{2} + k$, the minimum is at $(h,\, k)$.

Minimum point: $(1,\, -4)$

Axis of symmetry

The vertical line through the minimum is the axis of symmetry.

Axis of symmetry: $x = 1$

Axis: $x = 1$

You try

Find the minimum point of $f(x) = x^{2} + 4x + 1$.

Half the coefficient of $x$ is $2$. So aim for $(x+2)^{2}$.

$y = (x^{2} + 4x + 4) + 1 - 4$

$y = (x + 2)^{2} - 3$

Minimum at $(-2,\, -3)$

$(-2,\, -3)$

You try

What is the axis of symmetry of $g(x) = x^{2} - 6x + 11$?

Complete the square. Axis goes through the minimum.

$g(x) = (x - 3)^{2} + 11 - 9$

$g(x) = (x - 3)^{2} + 2$

Minimum at $(3, 2)$, so axis is the vertical line through $x = 3$.

$x = 3$

Section 4 of 12

Increasing & inequalities from a graph

Still working with $f(x) = x^{2} - 2x - 3$, minimum at $(1,\,-4)$.

i For what $x$ is $f(x)$ increasing?

A parabola is decreasing on the left of the minimum and increasing on the right.

Right of the vertex: $f$ is going up

$f$ is increasing for $x \ge 1$

ii For what $x$ is $f(x) \ge 0$?

$f(x) \ge 0$ means the curve is on or above the $x$-axis. From the graph, this is the parts outside the roots $-1$ and $3$.

Curve is at or above the $x$-axis outside the roots

$-2 \le x \le -1$ or $3 \le x \le 4$

Without the restricted domain, you'd just write $x \le -1$ or $x \ge 3$.

iii For what $x$ is $f(x) \ge 3$?

From section 2, $f(x) = 3$ at $x \approx -1.65$ and $x \approx 3.65$. The curve is above $y = 3$ outside those values.

Highlighted pieces show where $f \ge 3$ within the domain

$x \le -1.65$ or $x \ge 3.65$

You try

For the same parabola, for what $x$ is $f(x) < 0$?

$f(x) < 0$ means below the $x$-axis. That's the bit between the roots.

The curve is below the $x$-axis between $x = -1$ and $x = 3$.

Strict inequality so the endpoints are not included.

$-1 < x < 3$

Section 5 of 12

Two graphs together

Draw $f(x) = x^{2} - 2x$ and $g(x) = x$ on $-1 \le x \le 3$.

Quick table for $f(x) = x^{2} - 2x$:

$f(-1) = 1 + 2 = 3,\;\; f(0) = 0,\;\; f(1) = -1,\;\; f(2) = 0,\;\; f(3) = 3$

For $g(x) = x$, the points are simply $(-1,-1),\, (0,0),\, (1,1),\, (2,2),\, (3,3)$.

$f(x) = x^{2} - 2x$ (red) · $g(x) = x$ (blue)

i Find $x$ where $f(x) = g(x)$

"Meet" = intersection. From the graph the two cross at $(0,0)$ and $(3,3)$.

$x = 0,\; x = 3$

Check algebraically by solving $f(x) = g(x)$:

$x^{2} - 2x = x$

$x^{2} - 3x = 0$

$x(x - 3) = 0$

$x = 0,\; x = 3$ ✓

ii Find $x$ where $g(x) \ge f(x)$

"Where is blue at or above red?" — between the intersection points.

Shaded: where the line sits at or above the parabola

$0 \le x \le 3$

You try

For the same $f$ and $g$, for what $x$ is $f(x) > g(x)$ on the domain $-1 \le x \le 3$?

"Where is red above blue?" — the bit outside the meeting points (within the domain).

Red is above blue from $x = -1$ up to (but not including) $x = 0$.

It is not above blue again on this domain.

$-1 \le x < 0$

Section 6 of 12

The exponential graph

Draw $f(x) = 2^{x}$ on $-2 \le x \le 4$.

Quick table:

$f(-2) = 2^{-2} = \dfrac{1}{4} = 0.25$

$f(-1) = 2^{-1} = 0.5$

$f(0) = 2^{0} = 1$

$f(1) = 2,\; f(2) = 4,\; f(3) = 8,\; f(4) = 16$

$f(x) = 2^{x}$ — passes through $(0,1)$; never touches the $x$-axis

Key features: the curve never reaches zero on the left (the $x$-axis is a horizontal asymptote), and shoots up steeply on the right.

You try

For $f(x) = 2^{x}$, find $f(5)$ and $f(-3)$.

A negative power means a fraction: $2^{-3} = \dfrac{1}{2^{3}}$.

$f(5) = 2^{5} = 32$

$f(-3) = 2^{-3} = \dfrac{1}{2^{3}} = \dfrac{1}{8}$

$f(5) = 32,\; f(-3) = \dfrac{1}{8}$

$32$ and $\dfrac{1}{8}$

Section 7 of 12

Composition of functions

A composition is one function inside another — you apply one rule, then feed the result into the next.

Read it from the inside out · Must learn

$f \circ g(x) \;=\; f\bigl(g(x)\bigr)$ — do $g$ first, then $f$.

Given $f(x) = 3x + 1$ and $g(x) = 5x - 2$, find:

i $f \circ g(2)$

Do $g(2)$ first, then put the answer into $f$.

$g(2) = 5(2) - 2 = 8$

$f(8) = 3(8) + 1 = 25$

$f \circ g(2) = 25$

ii $g \circ f(2)$ (also written $gf(2)$)

Now do $f(2)$ first, then put it into $g$.

$f(2) = 3(2) + 1 = 7$

$g(7) = 5(7) - 2 = 33$

$g \circ f(2) = 33$

Order matters

In general $f \circ g(x)$ ≠ $g \circ f(x)$

You try

If $f(x) = 2x + 3$ and $g(x) = x - 4$, find $f \circ g(5)$.

Inside out: $g$ first.

$g(5) = 5 - 4 = 1$

$f(1) = 2(1) + 3 = 5$

$f \circ g(5) = 5$

$5$

You try

For the same $f$ and $g$, find $g \circ f(5)$.

This time $f$ first.

$f(5) = 2(5) + 3 = 13$

$g(13) = 13 - 4 = 9$

$g \circ f(5) = 9$

$9$

Section 8 of 12

Composition with expressions

Same idea, but keep $x$ in there instead of a number.

Given $f(x) = 5x - 1$ and $g(x) = x^{2} + 3$, find:

i $g \circ f(x)$

$g$ acts on $f(x) = 5x - 1$:

$g(5x - 1) = (5x - 1)^{2} + 3$

$= 25x^{2} - 10x + 1 + 3$

$= 25x^{2} - 10x + 4$

ii $f \circ g(x)$

$f$ acts on $g(x) = x^{2} + 3$:

$f(x^{2} + 3) = 5(x^{2} + 3) - 1$

$= 5x^{2} + 15 - 1$

$= 5x^{2} + 14$

iii $f^{2}(x)$

$f^{2}(x)$ means $f \circ f(x)$ — apply $f$ twice.

$f^{2}(x) = f(f(x)) = f(5x - 1)$

$= 5(5x - 1) - 1$

$= 25x - 5 - 1$

$= 25x - 6$

Confirm: $f^{2}(x)$ is the rule for "apply $f$ twice in a row". It is not $[f(x)]^{2}$.

You try

If $f(x) = 2x + 1$ and $g(x) = x^{2}$, find $g \circ f(x)$ in fully simplified form.

Put $f(x) = 2x+1$ inside $g$. Then expand.

$g(2x + 1) = (2x + 1)^{2}$

$= 4x^{2} + 4x + 1$

$g \circ f(x) = 4x^{2} + 4x + 1$

$4x^{2} + 4x + 1$

You try

For $h(x) = 3x - 4$, find $h^{2}(x)$.

$h^{2}(x) = h(h(x))$ — feed $3x-4$ back into $h$.

$h^{2}(x) = h(3x - 4)$

$= 3(3x - 4) - 4$

$= 9x - 12 - 4$

$= 9x - 16$

$9x - 16$

Section 9 of 12

Inverse functions: couples & arrows

The inverse of $f(x)$ is the rule that undoes it. We write it $f^{-1}(x)$.

Given $f(x) = 2x + 1$ on $-1 \le x \le 2$:

i Write down the couples for $f(x)$

$f(-1) = -1,\;\; f(0) = 1,\;\; f(1) = 3,\;\; f(2) = 5$

$(-1,\,-1),\;\; (0,\,1),\;\; (1,\,3),\;\; (2,\,5)$

ii Write down the couples for $f^{-1}(x)$

Just swap $x$ and $y$ in every couple.

$(-1,\,-1),\;\; (1,\,0),\;\; (3,\,1),\;\; (5,\,2)$

$f$ sends each input to its output. $f^{-1}$ runs the same arrows backwards.

Two things to remember · Must learn

Only a bijective function has an inverse (one-to-one and onto).

$f^{-1}(x)$ ≠ $\dfrac{1}{f(x)}$. The $-1$ means "inverse of", not a negative index.

Section 10 of 12

Inverse on the Cartesian plane

Plot $f(x)$ and $f^{-1}(x)$ on the same axes. Something beautiful happens.

$f$ and $f^{-1}$ are mirror images of each other in the line $y = x$.

Geometric fact · Must learn

$f^{-1}(x)$ is the image of $f(x)$ under axial symmetry in the line $y = x$.

A story to anchor it

Pays €10 an hour.

"How much do I earn in $5$ hours?" — that's $f(x)$: give the $x$, find $y$.

"How long to earn €$80$?" — that's $f^{-1}(x)$: give the $y$, find $x$.

Section 11 of 12

Finding the inverse: linear & rational

Method · 3 steps

1. Write $y = f(x)$.

2. Solve for $x$ in terms of $y$.

3. Swap $x$ and $y$, and that's $f^{-1}(x)$.

i Find $f^{-1}(x)$ for $f(x) = 2x + 1$

$y = 2x + 1$

$y - 1 = 2x$

$\dfrac{y - 1}{2} = x$

$f^{-1}(x) = \dfrac{x - 1}{2}$

ii Find $g^{-1}(x)$ for $g(x) = \dfrac{5x - 1}{3}$

$y = \dfrac{5x - 1}{3}$

$3y = 5x - 1$ (multiply both sides by 3)

$3y + 1 = 5x$

$\dfrac{3y + 1}{5} = x$

$g^{-1}(x) = \dfrac{3x + 1}{5}$

iii Find $f^{-1}(x)$ for $f(x) = \dfrac{3x - 1}{11}$

$y = \dfrac{3x - 1}{11}$

$11y = 3x - 1$

$11y + 1 = 3x$

$f^{-1}(x) = \dfrac{11x + 1}{3}$

You try

Find $f^{-1}(x)$ for $f(x) = 4x - 7$.

Set $y = 4x - 7$. Solve for $x$.

$y = 4x - 7$

$y + 7 = 4x$

$\dfrac{y + 7}{4} = x$

$f^{-1}(x) = \dfrac{x + 7}{4}$

You try

Find $g^{-1}(x)$ for $g(x) = \dfrac{2x + 5}{7}$.

Multiply both sides by $7$ first to clear the fraction.

$y = \dfrac{2x + 5}{7}$

$7y = 2x + 5$

$7y - 5 = 2x$

$\dfrac{7y - 5}{2} = x$

$g^{-1}(x) = \dfrac{7x - 5}{2}$

Section 12 of 12

Finding the inverse: quadratic

A quadratic is not bijective on $\mathbb{R}$ — so it doesn't have an inverse unless we restrict the domain. We'll be told what side of the parabola to take.

Find the inverse of $f(x) = x^{2} - 6x + 1$, given the domain $x \ge 3$. State the domain of the inverse.

First, a tempting wrong move — don't do this:

$y - 1 = x^{2} - 6x = x(x - 6)$ ✗ (can't split into $y - 1 = x$ and $y - 1 = x - 6$ — that's not how factors work)

Right move: complete the square first.

$y = x^{2} - 6x + 1$

$y = (x^{2} - 6x + 9) + 1 - 9$

$y = (x - 3)^{2} - 8$ (min at $(3,\,-8)$)

Now isolate $x$:

$y + 8 = (x - 3)^{2}$

$\sqrt{y + 8} = x - 3$ (positive root since $x \ge 3$)

$\sqrt{y + 8} + 3 = x$

$f^{-1}(x) = \sqrt{x + 8} + 3$

Domain of the inverse: the range of $f$ becomes the domain of $f^{-1}$. The minimum of $f$ is $-8$, so $f \ge -8$.

Domain of $f^{-1}$: $x \ge -8$

Find $g^{-1}(x)$ for $g(x) = x^{2} + 8x - 5$

$y = x^{2} + 8x + 16 - 5 - 16$

$y = (x + 4)^{2} - 21$ (min at $(-4,\,-21)$; take domain $x \ge -4$)

$y + 21 = (x + 4)^{2}$

$\sqrt{y + 21} = x + 4$

$\sqrt{y + 21} - 4 = x$

$g^{-1}(x) = \sqrt{x + 21} - 4$ · domain $x \ge -21$

You try

Find $f^{-1}(x)$ for $f(x) = x^{2} - 4x + 1$, given the domain $x \ge 2$. State the domain of the inverse.

Complete the square: half of $-4$ is $-2$, so add and subtract $4$.

$y = x^{2} - 4x + 4 + 1 - 4$

$y = (x - 2)^{2} - 3$ (min at $(2,\,-3)$)

$y + 3 = (x - 2)^{2}$

$\sqrt{y + 3} = x - 2$

$\sqrt{y + 3} + 2 = x$

$f^{-1}(x) = \sqrt{x + 3} + 2$, domain $x \ge -3$

$f^{-1}(x) = \sqrt{x + 3} + 2$, $x \ge -3$

You try

Find $f^{-1}(x)$ for $f(x) = x^{2} + 10x + 7$, given $x \ge -5$.

Half of $10$ is $5$; add and subtract $25$.

$y = x^{2} + 10x + 25 + 7 - 25$

$y = (x + 5)^{2} - 18$

$y + 18 = (x + 5)^{2}$

$\sqrt{y + 18} = x + 5$

$\sqrt{y + 18} - 5 = x$

$f^{-1}(x) = \sqrt{x + 18} - 5$, $x \ge -18$

That's Functions · Part 2.

Graphs, composition, and inverse. Every parabola question came back to the same handful of ideas: roots are at $y = 0$, the minimum drops out of completing the square, and the inverse is the mirror in $y = x$.