Sums and Products of Functions Revision Notes for VCE SSCE Mathematical Methods

Sums and Products of Functions

Combining functions

When working with functions, we can create new functions by adding or multiplying them together. This is useful for analyzing complex relationships and for sketching graphs.

Sum of functions

If we have two functions $f$ and $g$ , we can add them to create a new function called f + g. To find the value of this new function at any point $x$ , we simply add together the values of $f(x)$ and $g(x)$ .

Formal definition:

(f + g)(x) = f(x) + g(x)

Product of functions

Similarly, we can multiply two functions to create a product function $fg$ . At any point $x$ , we multiply the values $f(x)$ and $g(x)$ together.

Formal definition:

(fg)(x) = f(x) \cdot g(x)

Domain considerations

An important aspect of combining functions is determining where the new function is defined. Both $f(x)$ and $g(x)$ must be defined for the sum or product to exist.

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The domain of $f + g$ and the domain of $fg$ is the intersection of the individual domains. This means both functions must be defined at that $x$ -value for the combination to exist.

\text{dom}(f + g) = \text{dom } f \cap \text{dom } g

\text{dom}(fg) = \text{dom } f \cap \text{dom } g

The symbol $\cap$ means "intersection" - the values that belong to both domains.

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Worked Example: Combining Square Root Functions

Let's say we have $f(x) = \sqrt{x - 2}$ for all $x \geq 2$ and $g(x) = \sqrt{4 - x}$ for all $x \leq 4$ .

Step 1: Find the domain

First, we need to find where both functions are defined. The function $f$ requires $x \geq 2$ and $g$ requires $x \leq 4$ . The intersection of these domains is $[2, 4]$ .

Step 2: Finding the sum

(f + g)(x) = f(x) + g(x) = \sqrt{x - 2} + \sqrt{4 - x}

The domain is $[2, 4]$ .

To evaluate at $x = 3$ :

(f + g)(3) = \sqrt{3 - 2} + \sqrt{4 - 3} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2

Step 3: Finding the product

(fg)(x) = f(x) \cdot g(x) = \sqrt{x - 2} \cdot \sqrt{4 - x} = \sqrt{(x - 2)(4 - x)}

The domain is $[2, 4]$ .

To evaluate at $x = 3$ :

(fg)(3) = \sqrt{(3 - 2)(4 - 3)} = \sqrt{1 \times 1} = 1

Addition of ordinates

Addition of ordinates is a powerful graphical technique for sketching the sum of two functions. An ordinate is simply the y-value of a point on a graph. By adding the $y$ -values of two functions at each $x$ -value, we can construct the graph of $(f + g)(x)$ .

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This technique works well with other graphing methods like finding intercepts, stationary points, and asymptotes. It's particularly useful when you can't easily simplify the algebraic form of the sum.

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Worked Example: Addition of Ordinates with Linear Functions

Let's sketch $f(x) = x + 1$ and $g(x) = 3 - 2x$ , then use addition of ordinates to sketch $(f + g)(x)$ .

Step 1: Find the rule for the sum

(f + g)(x) = f(x) + g(x)

(f + g)(x) = (x + 1) + (3 - 2x)

(f + g)(x) = 4 - x

Step 2: Verify using ordinates

At any point, we can verify that the ordinate of $(f + g)(x)$ equals the sum of the ordinates of $f(x)$ and $g(x)$ .

For example, at $x = 2$ :

$f(2) = 2 + 1 = 3$
$g(2) = 3 - 2(2) = -1$
$(f + g)(2) = 3 + (-1) = 2$ ✓

Step 3: Create a table of values

We can create a table of values to help us plot points:

$x$	$f(x)$	$g(x)$	$(f + g)(x)$
$-1$	$0$	$5$	$5$
$0$	$1$	$3$	$4$
$\frac{3}{2}$	$\frac{5}{2}$	$0$	$\frac{5}{2}$
$2$	$3$	$-1$	$2$

Notice on the graph how at each $x$ -value, the height of the red line $(f + g)(x)$ equals the sum of the heights of the other two lines.

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Worked Example: Addition of Ordinates with Curved Functions

Now let's sketch $y = (f + g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x$ .

Step 1: Find the sum

(f + g)(x) = \sqrt{x} + x

Step 2: Determine the domain

Since both functions require $x \geq 0$ , the domain is $[0, \infty)$ .

Notice that the graph of $(f + g)(x)$ lies above both individual graphs since we're adding two positive functions together.

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Worked Example: Subtraction Using Addition of Ordinates

We can use the same technique for subtraction. Let's sketch $y = (f - g)(x)$ where $f(x) = x^2$ and $g(x) = \sqrt{x}$ .

Step 1: Find the difference

(f - g)(x) = x^2 - \sqrt{x}

This is equivalent to adding $f(x)$ and $-g(x)$ (the negative of $g$ ).

Step 2: Determine the domain

The domain is $[0, \infty)$ since both functions must be defined.

The graph shows $f(x)$ in blue, $-g(x)$ in yellow-orange, and $(f - g)(x)$ in red. At each $x$ -value, we add the ordinates of $f$ and $-g$ to get the ordinate of $(f - g)$ .

Key points for sketching sums

When sketching $y = (f + g)(x)$ using addition of ordinates, keep these important observations in mind:

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Critical Checkpoints When Sketching:

When f(x) = 0, then $(f + g)(x) = g(x)$ - the sum graph touches the $g$ graph
When g(x) = 0, then $(f + g)(x) = f(x)$ - the sum graph touches the $f$ graph
If both f(x) and g(x) are positive, then $(f + g)(x) > g(x)$ and $(f + g)(x) > f(x)$ - the sum is above both graphs
If both f(x) and g(x) are negative, then $(f + g)(x) < g(x)$ and $(f + g)(x) < f(x)$ - the sum is below both graphs
If f(x) is positive and g(x) is negative, then $g(x) < (f + g)(x) < f(x)$ - the sum is between the two graphs
Look for values of $x$ where f(x) + g(x) = 0 - these are $x$ -intercepts of the sum

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Key Points to Remember:

The sum of functions is defined by $(f + g)(x) = f(x) + g(x)$
The product of functions is defined by $(fg)(x) = f(x) \cdot g(x)$
The domain of both sum and product is the intersection of the individual domains: $\text{dom}(f + g) = \text{dom}(fg) = \text{dom } f \cap \text{dom } g$
Addition of ordinates is a graphical technique where you add the y-values of two functions at each $x$ -value to sketch their sum
Key checkpoints: where one function equals zero, the sum equals the other function

Sums and Products of Functions (VCE SSCE Mathematical Methods): Revision Notes