Families of Functions and Solving Literal Equations Revision Notes for VCE SSCE Mathematical Methods

Families of Functions and Solving Literal Equations

Introduction to families of functions

A family of functions consists of related functions that share the same general form but differ by one or more parameters. Parameters are constants (typically represented by letters such as $m$ , $k$ , $a$ , $b$ , or $c$ ) that can take various values within a specified domain.

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Using parameters allows us to describe and analyze general properties that apply to all functions within a family, rather than examining each function individually. This powerful approach enables us to understand entire collections of related functions at once.

Examples of function families

Here are several common families of functions:

Linear through origin: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = mx$ where $m \in \mathbb{R}$
Cubic: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = ax^3$ where $a \in \mathbb{R} \setminus \{0\}$
Linear with fixed y-intercept: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = mx + 2$ where $m \in \mathbb{R}^+$
Exponential: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = ke^{mx}$ where $m \in \mathbb{R} \setminus \{0\}$ and $k \in \mathbb{R} \setminus \{0\}$

Each of these families represents infinitely many functions, with each specific value of the parameter(s) producing a different member of the family.

Working with linear function families

Let's explore how to work with a family of linear functions where the y-intercept is fixed but the gradient varies.

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Worked Example: Linear Function Family $f(x) = mx + 2$ where $m \in \mathbb{R}^+$

a) Find the x-axis intercept of the graph of $y = f(x)$

The x-axis intercept occurs when $y = 0$ :

mx + 2 = 0

mx = -2

x = -\frac{2}{m}

The x-axis intercept is $-\frac{2}{m}$ .

Note that since $m > 0$ , the x-intercept is always negative for this family of functions.

b) For which values of $m$ is the x-axis intercept less than $-2$ ?

We need to solve the inequality:

-\frac{2}{m} < -2

When solving inequalities with fractions, we need to be careful about the sign. Since we know $m > 0$ , multiplying both sides by $m$ preserves the inequality direction:

\frac{2}{m} > 2

2 > 2m

m < 1

Therefore, the x-axis intercept is less than $-2$ when $0 < m < 1$ .

c) Find the inverse function of $f$

To find the inverse, we swap $x$ and $y$ , then solve for $y$ :

Start with: $y = mx + 2$

Swap variables: $x = my + 2$

Solve for $y$ :

my = x - 2

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

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

Therefore, $f^{-1}(x) = \frac{1}{m}x - \frac{2}{m}$ .

The domain of $f^{-1}$ is $\mathbb{R}$ .

d) Find the equation of the line perpendicular to the graph of $y = f(x)$ at the point $(0, 2)$

The gradient of $y = mx + 2$ is $m$ .

For perpendicular lines, the product of their gradients equals $-1$ :

m \times m_{\perp} = -1

m_{\perp} = -\frac{1}{m}

Using the point-gradient form with point $(0, 2)$ and gradient $-\frac{1}{m}$ :

y - 2 = -\frac{1}{m}(x - 0)

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

This perpendicular line has the same y-intercept but a negative reciprocal gradient.

chatImportant

Common Mistake to Avoid: When solving inequalities involving parameters, always consider the sign of the parameter before multiplying or dividing both sides. If the parameter could be negative, the inequality direction would reverse!

Working with quadratic function families

When working with families of quadratic functions, we can express unknown coefficients in terms of a known parameter. This approach is particularly useful when partial information about the function is given.

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Worked Example: Quadratic Function Through Given Points

The graph of a quadratic function passes through the points $(1, 6)$ and $(2, 4)$ . Find the coefficients of the quadratic rule in terms of $c$ , the y-axis intercept of the graph.

Let $f(x) = ax^2 + bx + c$ represent the general quadratic in this family.

Since the graph passes through $(1, 6)$ and $(2, 4)$ :

$f(1) = 6$ gives: $a + b + c = 6$

$f(2) = 4$ gives: $4a + 2b + c = 4$

We now have two equations with three unknowns. We can solve for $a$ and $b$ in terms of $c$ :

From the first equation: $a + b = 6 - c$ ... (1)

From the second equation: $4a + 2b = 4 - c$ ... (2)

Multiply equation (1) by 2: $2a + 2b = 12 - 2c$ ... (3)

Subtract equation (3) from equation (2):

2a = (4 - c) - (12 - 2c)

2a = 4 - c - 12 + 2c

2a = c - 8

a = \frac{c - 8}{2}

Substitute back into equation (1):

\frac{c - 8}{2} + b = 6 - c

b = 6 - c - \frac{c - 8}{2}

b = \frac{12 - 2c - c + 8}{2}

b = \frac{20 - 3c}{2}

The equation of the quadratic in terms of $c$ is:

y = \left(\frac{c - 8}{2}\right)x^2 + \left(\frac{20 - 3c}{2}\right)x + c

Solving literal equations

A literal equation is an equation containing multiple variables and parameters. When solving literal equations involving exponential and logarithmic functions, we use the same fundamental principles as with numerical equations.

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Worked Example: Solving Literal Equations with Exponentials and Logarithms

Solve each of the following for $x$ . All constants are positive reals.

a) $ae^{bx} - c = 0$

ae^{bx} = c

e^{bx} = \frac{c}{a}

Taking the natural logarithm of both sides:

bx = \log_e\left(\frac{c}{a}\right)

x = \frac{1}{b}\log_e\left(\frac{c}{a}\right)

b) $\log_e(x - a) = b$

Convert from logarithmic to exponential form:

x - a = e^b

x = e^b + a

c) $\log_e(cx - a) = 1$

Convert from logarithmic to exponential form (remembering that $e^1 = e$ ):

cx - a = e

cx = a + e

x = \frac{a + e}{c}

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Key Techniques for Literal Equations:

When solving literal equations:

Isolate the term containing $x$ first before applying inverse operations
Use inverse operations strategically: exponential form ↔ logarithmic form
Be careful when multiplying or dividing inequalities by parameters (sign matters!)
Consider any restrictions on parameter values (e.g., positive, non-zero) to ensure your solution is valid

Transformations with parameters

Transformations of the plane can involve parameters, allowing us to describe families of transformations. This extends the concept of function families to geometric transformations.

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Worked Example: Transformation with Parameter

A transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ is defined by the rule:

$(x, y) \to (5x + 5, ky + 2)$

where $k$ is a non-zero real number.

a) Let $(x', y') = (5x + 5, ky + 2)$ . Find $x$ and $y$ in terms of $x'$ and $y'$ .

We have two equations:

5x + 5 = x'

ky + 2 = y'

Solving for $x$ :

5x = x' - 5

x = \frac{x' - 5}{5}

Solving for $y$ :

ky = y' - 2

y = \frac{y' - 2}{k}

b) Find the image of the curve with equation $y = \frac{1}{x}$ under this transformation.

From part a), we know:

x = \frac{x' - 5}{5} \text{ and } y = \frac{y' - 2}{k}

Substituting into $y = \frac{1}{x}$ :

\frac{y' - 2}{k} = \frac{1}{\frac{x' - 5}{5}}

\frac{y' - 2}{k} = \frac{5}{x' - 5}

Solving for $y'$ :

y' - 2 = \frac{5k}{x' - 5}

y' = \frac{5k}{x' - 5} + 2

Since we typically write the image equation using $(x, y)$ rather than $(x', y')$ :

y = \frac{5k}{x - 5} + 2

c) Find the value of $k$ if the image passes through the origin.

If the image passes through the origin, then the point $(0, 0)$ satisfies the image equation:

0 = \frac{5k}{0 - 5} + 2

0 = \frac{5k}{-5} + 2

0 = -k + 2

k = 2

bookmarkSummary

Key Points to Remember:

Families of functions use parameters to describe collections of related functions with shared properties
When solving inequalities involving parameters, pay careful attention to whether multiplying or dividing by the parameter reverses the inequality sign (this depends on whether the parameter is positive or negative)
To find inverse functions with parameters, swap $x$ and $y$ , then solve for $y$ in terms of $x$ and the parameter
The gradient of a line perpendicular to a line with gradient $m$ is $-\frac{1}{m}$ (their product equals $-1$ )
When solving literal equations with exponentials and logarithms, use inverse operations: exponential form ↔ logarithmic form
For transformations involving parameters, finding the inverse transformation allows you to express the image of a curve

Families of Functions and Solving Literal Equations (VCE SSCE Mathematical Methods): Revision Notes