Graph Of Functions

Is the set of all possible values of x for which the corresponding value of y is known.

Example

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

Is the set of all possible values of y for which the corresponding value of x is known

Example

$y = 2x$

Relation as a mapping

In x-y plane (ordered pair)

Is the mapping a single element from domain into range?

Not a function

The following are some types of function:

1. Constant Function

$f(x) = c$

2. Linear Function

$f(x) = ax + b$

3. Quadratic Function

$f(x) = ax^2 + bx + c$

4. Absolute Value Function

$f(x) = |x|$

5. Rational Function

$f(x) = \frac{ax + b}{cx + d}$

Sketching the function

Suppose given the function:

If $f(x) = y$

Given the function:

Suppose $y = x$ for which $x > 0$

$y = 1 - x$

$y = 1 + x$ for $x < 0$

Given the function $f(x)$:

Solution

Suppose that $f(x) = y$

$y = x^2 - 1$, for $x > 0$

$y = x + 1$ for $x < 0$

$y = 3$ for $x > 5$

Sometimes referred to as compound function, are linear functions whose variables have a special relationship under certain conditions that make their graphs break in intervals (look like steps). To understand the concept, let us look at the following example.

The cost of shaving the hair of different age-groups in a central salon are as follows:

1. Shaving the group aged ten to twenty years costs Tsh 2,000/=
2. The group aging between twenty and thirty (exclusive) costs Tsh 4,000/=
3. The group aging thirty and above costs Tsh 6,000/=

From the above information, provide domain and range.

Solution

If we let $x$ be the ages and $f(x)$ be the costs, then we can interpret this problem as a step function defined by:

The domain of this function is a set of real numbers such that $x \geq 10$.

The range of this function is $\{2000, 4000, 6000\}$.

A quadratic function is a polynomial of the second degree.

It is a function of the general form $ax^2 + bx + c$

Where $a$, $b$, and $c$ are real numbers and $a \neq 0$

Example

Draw the graph of the function:

1. $f(x) = x^2 - 1$
2. $f(x) = -x^2 - 1$

Solution

Table of Values

(i) $f(x) = x^2 - 1$

Its graph

(ii) $f(x) = -x^2 - 1$

Table of Values

Its graph

When the polynomial function is reduced to the third degree, a cubic function is obtained.

The cubic function takes the general form $f(x) = ax^3 + bx^2 + cx + d$

Where $a$, $b$, $c$, and $d$ are real numbers and $a \neq 0$

Example

Draw the graph of the following function:

$f(x) = x^3 - 9x$

• The intercepts are points $(-3, 0)$, $(0, 0)$, $(3, 0)$
• There are two turning points; the maximum i.e. $(-2, 16)$ and the minimum i.e. $(2, -16)$
• The domain is the set of all real numbers
• The range is the set of all real numbers

For the turning point, let us consider the function $f(x) = ax^2 + bx + c$. The function $f$ may be expressed in the form $a[g(x)] + k$

Where $g(x)$ is another function in $x$ and $k$ is a constant as follows.

$f(x) = ax^2 + bx + c$

Factorizing out the constant $a$:

Recall the condition that:

1. If $a > 0$, $f$ has a minimum value $K$, which implies that $f(x) = y = \frac{4ac - b^2}{4a}$ when $x = -\frac{b}{2a}$. Hence, the minimum turning point is $\left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)$ when $a$ is positive.
2. If $a < 0$, $f$ has a maximum value $K = \frac{4ac - b^2}{4a}$ which occurs when $x = -\frac{b}{2a}$. Therefore, the maximum turning point on the graph of $f$ is $\left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right)$ when $a$ is negative.

Example

Sketch the graph of $f(x) = x^2 + 2x - 8$, determine the turning point and the intercepts.

Solution

$x^2 + 2x - 8 = 0$

Solving we get:

$(x + 4)(x - 2) = 0$

$x + 4 = 0$, $x - 2 = 0$

$x = -4$, $x = 2$ which are the x-intercepts.

• The y-intercept is $-8$.

• To obtain the turning points, equate $x^2 + 2x - 8 = 0$ to $ax^2 + bx + c = 0$, so that by comparison we get:

$a = 1$, $b = 2$, $c = -8$

Thus:

Turning point $= \left( -\frac{2}{2(1)}, \frac{4(1)(-8) - (2)^2}{4(1)} \right)$

$= (-1, -9)$

There are lines which the curve does not touch. There are two types of asymptotes:

1. Vertical asymptotes
2. Horizontal asymptotes

Is the line where the function approaches infinity as x approaches a certain value.

Is the line where the function approaches a constant value as x approaches infinity.

Example 1

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

Let $f(x) = y$

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

Vertical Asymptote (V.A.)

For V.A., we set the denominator equal to zero:

$x - 2 = 0$

$\text{V.A.} = 2$

Horizontal Asymptote (H.A.)

To find H.A., divide numerator and denominator by $x$:

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

As $x \to \infty$, $\frac{1}{x} \to 0$

$y = \frac{2}{1} = 2$

$\text{H.A.} = 2$

Intercepts:

Example 2

Sketch the function:

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

Suppose $f(x) = y$

$y = \frac{x + 1}{x^2 + 1}$

Vertical Asymptotes

For V.A., set denominator equal to zero:

$x^2 + 1 = 0$

$x^2 = -1$

No real vertical asymptotes.

Horizontal Asymptotes

To find H.A., divide numerator and denominator by $x^2$:

$y = \frac{\frac{1}{x} + \frac{1}{x^2}}{1 + \frac{1}{x^2}}$

As $x \to \infty$, $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$

$y = \frac{0}{1} = 0$

$\text{H.A.} = 0$

Intercepts

When $x = 0$, $y = 1$

When $y = 0$, $x = -1$

Practice Problems

Sketch the following functions:

(a) $f(x) = x - \frac{1}{x}$

(b) $f(x) = \frac{1 - 3x}{x - 2}$

(c) $f(x) = \frac{2x + 1}{3x + 2}$

(d) $f(x) = \frac{4}{x^2 - 5x + 4}$

Solution (a): $f(x) = x - \frac{1}{x}$

Let $f(x) = y$

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

Vertical Asymptote

For V.A., $\frac{1}{x} \to \infty$ as $x \to 0$

$\text{V.A.} = 0$

Horizontal Asymptotes

As $x \to \infty$, $\frac{1}{x} \to 0$

$y = x$ (no horizontal asymptote, but oblique asymptote $y = x$)

Intercepts

When $x = 1$, $y = 0$

When $y = 0$, $x = \pm 1$

Solution (b): $f(x) = \frac{1 - 3x}{x - 2}$

Let $f(x) = y$

$y = \frac{1 - 3x}{x - 2}$

Vertical Asymptote

Set denominator to zero:

$x - 2 = 0$

$x = 2$

$\text{V.A.} = 2$

Horizontal Asymptote

Divide numerator and denominator by $x$:

$y = \frac{\frac{1}{x} - 3}{1 - \frac{2}{x}}$

As $x \to \infty$, $\frac{1}{x} \to 0$

$y = \frac{-3}{1} = -3$

$\text{H.A.} = -3$

Intercepts

When $x = 0$, $y = -\frac{1}{2}$

When $y = 0$, $x = \frac{1}{3}$

Solution (c): $f(x) = \frac{2x + 1}{3x + 2}$

Let $f(x) = y$

$y = \frac{2x + 1}{3x + 2}$

Vertical Asymptote

Set denominator to zero:

$3x + 2 = 0$

$x = -\frac{2}{3}$

$\text{V.A.} = -\frac{2}{3}$

Horizontal Asymptote

Divide numerator and denominator by $x$:

$y = \frac{2 + \frac{1}{x}}{3 + \frac{2}{x}}$

As $x \to \infty$, $\frac{1}{x} \to 0$ and $\frac{2}{x} \to 0$

$y = \frac{2}{3}$

$\text{H.A.} = \frac{2}{3}$

Intercepts

When $x = 0$, $y = \frac{1}{2}$

When $y = 0$, $x = -\frac{1}{2}$

Solution (d): $f(x) = \frac{4}{x^2 - 5x + 4}$

Let $f(x) = y$

$y = \frac{4}{x^2 - 5x + 4}$

Vertical Asymptotes

Set denominator equal to zero:

$x^2 - 5x + 4 = 0$

$(x - 1)(x - 4) = 0$

$x = 1$ or $x = 4$

$\text{V.A.} = 1$ and $\text{V.A.} = 4$

Horizontal Asymptote

Divide numerator and denominator by $x^2$:

$y = \frac{\frac{4}{x^2}}{1 - \frac{5}{x} + \frac{4}{x^2}}$

As $x \to \infty$, $\frac{1}{x} \to 0$ and $\frac{1}{x^2} \to 0$

$y = \frac{0}{1} = 0$

$\text{H.A.} = 0$

Intercepts

When $x = 0$, $y = 1$

No x-intercepts (numerator is constant 4)

Example

Given the function:

$f(x) = \frac{(x - 5)(x - 1)}{(x + 1)(x - 3)}$

(i) State the range of the function (the domain is the set of real values)

(ii) Sketch the function $f(x)$

Solution

Let $f(x) = y$

$y = \frac{(x - 5)(x - 1)}{(x + 1)(x - 3)}$

$y = \frac{x^2 - 6x + 5}{x^2 - 2x - 3}$

To find the range, we solve for $x$ in terms of $y$:

For real values of $x$, the discriminant must be non-negative:

$b^2 - 4ac \geq 0$

$(-2y + 6)^2 - 4(y - 1)(-(3y + 5)) \geq 0$

$4y^2 - 24y + 36 + 4(y - 1)(3y + 5) \geq 0$

$4y^2 - 24y + 36 + 12y^2 + 20y - 12y - 20 \geq 0$

$16y^2 - 16y + 16 \geq 0$

$y^2 - y + 1 \geq 0$

$\left(y - \frac{1}{2}\right)^2 + \frac{3}{4} \geq 0$

This is always true for all real $y$.

Therefore, $y$ has no restriction: it can be any real value.

Range: All real numbers

Sketch of $f(x) = \frac{(x - 5)(x - 1)}{(x + 1)(x - 3)}$

Vertical Asymptotes

Set denominator equal to zero:

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

$x + 1 = 0$ or $x - 3 = 0$

$\text{V.A.} = -1$, $\text{V.A.} = 3$

Horizontal Asymptote

$y = \frac{x^2 - 6x + 5}{x^2 - 2x - 3}$

Divide by $x^2$:

$y = \frac{1 - \frac{6}{x} + \frac{5}{x^2}}{1 - \frac{2}{x} - \frac{3}{x^2}}$

As $x \to \infty$, $\frac{1}{x} \to 0$

$y = \frac{1}{1} = 1$

$\text{H.A.} = 1$

Intercepts

When $x = 0$, $y = \frac{(-5)(-1)}{(-1)(-3)} = \frac{5}{3}$

When $y = 0$, $x = 1$ or $x = 5$

Example

Show that there are only two values which $\frac{(2x - 1)(2x + 5)}{2x - 3}$ cannot lie between, and find them.

Approach

$y = \frac{(2x - 1)(2x + 5)}{2x - 3}$

Multiply both sides by $(2x - 3)$:

$y(2x - 3) = (2x - 1)(2x + 5)$

$2xy - 3y = 4x^2 + 10x - 2x - 5$

$2xy - 3y = 4x^2 + 8x - 5$

Rearrange to form quadratic in $x$:

$4x^2 + 8x - 2xy - 5 + 3y = 0$

$4x^2 + (8 - 2y)x + (3y - 5) = 0$

For real values of $x$, the discriminant must be non-negative:

$b^2 - 4ac \geq 0$

$(8 - 2y)^2 - 4(4)(3y - 5) \geq 0$

$64 - 32y + 4y^2 - 48y + 80 \geq 0$

$4y^2 - 80y + 144 \geq 0$

Divide by 4:

$y^2 - 20y + 36 \geq 0$

Factor:

$(y - 2)(y - 18) \geq 0$

This inequality holds when:

$y \leq 2$ or $y \geq 18$

Therefore, the function cannot lie between 2 and 18 (exclusive).

The values are 2 and 18.

Two functions $f$ and $g$ are said to be composite functions of $f \circ g = f(g(x))$

Note: Composite functions are not commutative in general.

Given $f(x) = x^2 + 1$ and $g(x) = 2x$.

Find:

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

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

Approach

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

$f(2x) = (2x)^2 + 1 = 4x^2 + 1$

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

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

$f \circ g \neq g \circ f$, hence the composite function is not commutative.

Given:

$f(x) = x^2 - 1$, $g(x) = 3x$ and $h(x) = \frac{2}{x}$

(i) $(f \circ g) \circ h$

(ii) $f \circ (g \circ h)$

Solution

$f \circ g = f(g(x)) = f(3x) = (3x)^2 - 1 = 9x^2 - 1$

$(f \circ g) \circ h = (f \circ g)(h(x)) = (f \circ g)\left(\frac{2}{x}\right) = 9\left(\frac{2}{x}\right)^2 - 1 = \frac{36}{x^2} - 1$

(ii) $g \circ h = g(h(x)) = g\left(\frac{2}{x}\right) = 3\left(\frac{2}{x}\right) = \frac{6}{x}$

$f \circ (g \circ h) = f\left(\frac{6}{x}\right) = \left(\frac{6}{x}\right)^2 - 1 = \frac{36}{x^2} - 1$

Since $(f \circ g) \circ h = f \circ (g \circ h)$, hence the composite function satisfies the associative property.

A function is a one-to-one function when a line parallel to the x-axis cuts the curve at most once. Equivalently, a line parallel to the y-axis also cuts the curve at most once.

The line parallel to the x-axis cuts the curve only once:

• An inverse function is one where each element from the domain matches exactly in the range, and conversely, each element from the range matches exactly with the domain.

Example 1

Given $f(x) = 2x - 1$

Find $f^{-1}(x)$

Approach

Let $f^{-1}(x) = g(x)$

$f(g(x)) = x$

$2g(x) - 1 = x$

$2g(x) = x + 1$

$g(x) = \frac{x + 1}{2}$

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

Example 2

Given $f(x) = x^2 + 1$

Find $f^{-1}(x)$

Approach

Let $f^{-1}(x) = g(x)$

$f(g(x)) = x$

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

$(g(x))^2 = x - 1$

$g(x) = \sqrt{x - 1}$ (for $x \geq 1$)

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

Example 3

Given:

Sketch:

(i) $f(x)$ – state its domain

(ii) $f^{-1}(x)$

Solution

(i) The domain of $f(x)$ is all real numbers, $\mathbb{R}$.

For $f(x) = x + 1$ (when $x < 0$):

Let $f^{-1}(x) = g(x)$

$f(g(x)) = x$

$g(x) + 1 = x$

$g(x) = x - 1$

For $f(x) = x^2$ (when $x \geq 0$):

Let $f^{-1}(x) = g(x)$

$f(g(x)) = x$

$(g(x))^2 = x$

$g(x) = \sqrt{x}$ (for $x \geq 0$)