Function Notation and Identities Revision Notes for VCE SSCE Mathematical Methods

Function Notation and Identities

Understanding function notation for familiar properties

Properties of functions that you have already studied can be expressed using function notation. This approach helps you see patterns more clearly and understand how different functions behave.

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Using function notation to express mathematical properties makes it easier to identify patterns and relationships between different types of functions. This notation is particularly powerful when working with logarithmic and exponential functions.

Logarithmic functions

The familiar rules for logarithms can be written in a more compact form using function notation.

Standard logarithm rules:

\log_e(x) + \log_e(y) = \log_e(xy)

\log_e(x) - \log_e(y) = \log_e\left(\frac{x}{y}\right)

If we write $f(x) = \log_e x$ , these rules become:

f(x) + f(y) = f(xy)

f(x) - f(y) = f\left(\frac{x}{y}\right)

This notation shows that adding logarithms corresponds to multiplying their arguments, whilst subtracting logarithms corresponds to dividing their arguments.

Exponential functions

Similarly, the rules for exponential functions can be expressed using function notation.

Standard exponential rules:

e^{x+y} = e^x \times e^y

e^{x-y} = \frac{e^x}{e^y}

If we write $f(x) = e^x$ , these rules become:

f(x + y) = f(x)f(y)

f(x - y) = \frac{f(x)}{f(y)}

This shows that adding exponents corresponds to multiplying exponential values, whilst subtracting exponents corresponds to dividing exponential values.

Worked example: linear function

For the function with rule $f(x) = 2x$ , we can show that $f(x + y) = f(x) + f(y)$ for all $x$ and $y$ .

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Worked Example: When the property holds

For $f(x) = 2x$ :

f(x + y) = 2(x + y)

= 2x + 2y

= f(x) + f(y)

This confirms that for $f(x) = 2x$ , the property $f(x + y) = f(x) + f(y)$ is true for all values of $x$ and $y$ .

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Worked Example: When the property doesn't hold

For the function with rule $f(x) = x + 2$ :

f(x + y) = (x + y) + 2

= (x + 2) + (y + 2) - 2

= f(x) + f(y) - 2

If $f(x) + f(y) - 2 = f(x) + f(y)$ , then $-2 = 0$ , which is a contradiction.

Therefore, $f(x + y) \neq f(x) + f(y)$ for this function.

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Not all functions satisfy the property $f(x + y) = f(x) + f(y)$ . The constant term in $f(x) = x + 2$ prevents this property from holding.

Worked example: reciprocal function

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Worked Example: Verifying the reciprocal function identity

If $f(x) = \frac{1}{x}$ , we can verify that $f(x) + f(y) = (x + y)f(xy)$ for all non-zero real numbers $x$ and $y$ .

Starting with the left side:

f(x) + f(y) = \frac{1}{x} + \frac{1}{y}

= \frac{y + x}{xy}

= (x + y) \times \frac{1}{xy}

= (x + y)f(xy)

This confirms the identity holds for all non-zero real numbers $x$ and $y$ .

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We must specify that $x$ and $y$ are non-zero because the function $f(x) = \frac{1}{x}$ is undefined at $x = 0$ . This is an important domain restriction to remember.

Worked example: trigonometric function

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Worked Example: Finding a counterexample for cosine

For the function $f(x) = \cos x$ , we can show with a specific example that $f(x + y) \neq f(x) + f(y)$ for some values of $x$ and $y$ .

Let's test with $x = 0$ and $y = \pi$ :

f(0 + \pi) = f(\pi) = -1

f(0) + f(\pi) = 1 + (-1) = 0

Since $-1 \neq 0$ , we have shown that $f(x + y) \neq f(x) + f(y)$ for these values.

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To show that a property doesn't hold for all values, you only need to find one counterexample. The values $x = 0$ and $y = \pi$ are often useful test values for trigonometric functions.

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Key Points to Remember:

Properties of functions can be expressed concisely using function notation
For logarithms: $f(x) + f(y) = f(xy)$ when $f(x) = \log_e x$
For exponentials: $f(x + y) = f(x)f(y)$ when $f(x) = e^x$
The property $f(x + y) = f(x) + f(y)$ does not hold for all functions
To disprove a property, find one counterexample with specific values

Function Notation and Identities (VCE SSCE Mathematical Methods): Revision Notes