# VCE Mathematical Methods/Printable version

VCE Mathematical Methods

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# Inverse functions

## Theory

### The Existence of Inverse Functions

A function has an inverse if and only if it is one-to-one: that is, if for each y-value there is only one corresponding x-value. To test whether or not a function is one-to-one, we can draw multiple horizontal lines through the graph of the function. If any of these horizontal lines intersects with the graph of the function more than once, then the function is NOT one-to-one.

For example, take the graph of the function $f:\mathbb {R} \to \mathbb {R} ,f(x)=x^{2}$ .

We can draw a horizontal line anywhere on the graph (except for the turning point at $x=0$ ) which will intersect with the graph twice. Therefore f is NOT a one-to-one function, and will NOT have an inverse function.

However, consider the function $g:\mathbb {R} \to \mathbb {R} ,g(x)=x^{3}$ .

This function IS one-to-one, and will therefore have an inverse function, which we label $g^{-1}\,$ .

### The rule of an inverse function

The rule of the inverse of a function f can be found by letting $y=f(x)$ , swapping the y and x variables, and re-arranging to make y the subject.

For example, consider g. To find the rule of $g^{-1}(x)$ , let $g(x)=y$  Now $y=x^{3}\,$ . Swap the x and y variables to get $x=y^{3}\,$ . Now take the cube root of each side to get $y={\sqrt[{3}]{x}}\,$ . Now let $y=g^{-1}(x)$  and we have:-

$g^{-1}(x)={\sqrt[{3}]{x}}$

### The domain and range of inverse functions

Because, in finding inverse functions, we are swapping the x and y values in each ordered pair, it's logical that the domain of a function is the range of the inverse; and the range of a function is the domain of the inverse.

Thus: ${\mbox{dom}}f={\mbox{ran}}f^{-1}{\mbox{ and ran}}f={\mbox{dom}}f^{-1}\,$

# Differentiation from First Principles

## Theory

### Formula

Given a function f, the rule of the derivative (sometimes called the "gradient") function is defined as $f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\,$ .

### Method

Remember that in order to evaluate a limit, we usually substitute the value given into the expression. However, with the above formula, substituting $h=0$  will result in a division by zero, which is mathematically impossible. Therefore,in order to make use of this formula, you need to substitute the rules $f(x+h)$  and $f(x)$ , then simplify to eliminate the fraction, and only then substitute $h=0$ . This is called differentiation from first principles.

For example:

Let $f:\mathbb {R} \to \mathbb {R} ,f(x)=2x$

Let us differentiate f from first principles.

{\begin{aligned}f'(x)&=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\\&=\lim _{h\to 0}{\frac {2(x+h)-2x}{h}}\\&=\lim _{h\to 0}{\frac {2x+2h-2x}{h}}\\&=\lim _{h\to 0}{\frac {2h}{h}}\\&=\lim _{h\to 0}2\\&=2\\\end{aligned}} .
Therefore, we can define the gradient function as $f':\mathbb {R} \to \mathbb {R} ,f'(x)=2$

## Exercises

Question One
Differentiate the following functions from first principles.
(a) $f(x)=4x$
(b) $f(x)=7x$
(c) $f(x)=2x+1$
(d) $f(x)=3x+3$

Question Two
Differentiate the following functions from first principles.
(a) $g(x)=x^{2}$
(b) $f(x)=5x^{2}$
(c) $f(x)=2x^{2}+3$
(d) $f(x)=(x+3)(x+4)$

# Exam One Practice One

## Instructions

Writing Time: 60 minutes

• Students are permitted to use: pencils, pens, highlighters, erasers, sharpeners, rulers, protractors, set-squares, aids for curve sketching
• Students are NOT permitted to use: blank sheets of paper, white-out, any type of technology
• Any diagrams used are NOT drawn to scale unless otherwise indicated
• Students must answer all the questions in the space provided
• In questions where more than one mark is available, appropriate working MUST be shown
• When instructed to use calculus, an appropriate derivative or anti-derivative MUST be shown

## Questions

### Question 1

(a) Given $e^{3x+1}-1=0\,$ , solve for x.

(b) If $f(x)=e^{3x+1}-1\,$  and $g(x)=e^{x}\,$  state the transformations required to change g into f

[1 + 2 = 3 marks]

### Question 2

Let $P(x)=x^{4}+2x^{3}-9x^{2}-2x+8\,$  and $Q(x)=x-1\,$ .

(a) Evaluate ${\frac {P(x)}{Q(x)}}$

(b) Hence factorise P(x) given that $P(2)=0$ .

(c) Hence sketch the graph of P

[2+2+2 = 6 marks]

### Question 3

Let $f:[0,\pi )\to \mathbb {R} ,f(x)=-cos(x)-x$ . Use calculus to find the co-ordinates of the stationary point.

[3 marks]

### Question 4

A garden path can be modelled with the equation $y=sin(2x)+1\,$  where $x\in [0,2\pi ]$ .

(a)Sketch the garden path over the domain specified.

(b) If the x-axis represents a fence, use calculus to determine the area between the path and the fence.

[2+2 = 4 marks]

### Question 5

State the equations of the tangent and the normal of the function $h:(-\infty ,-2)\cup (-2,\infty )\to \mathbb {R} ,h(x)=log_{e}(x+2)+3\,$  when $x=1\,$

[2 marks]

### Question 6

Shirley either eats lamingtons or a muesli bar for morning tea. If Shirley eats lamingtons one day, then the probability she will eat lamingtons the next day is 0.5. If Shirley eats a muesli bar one day, the probability that she will eat a muesli bar the next day is 0.3. If Shirley eats a muesli bar on Tuesday, what is the probability she will eat lamingtons on Thursday?

[2 marks]