Part 1

Substitute 2 for any place where there is an .

Part 2

Subbing in , ensuring to square the negative as well:

Part 3

Notice that for all the previous problems, any time a substitution is made, the substituted value is enclosed in brackets. This helps you keep track of what has changed, and will be easy to see if any additional operations need to be made, like expanding signs, powers or terms.

Replace every with the expression . This gives us:

Now just expand and simplify where possible.

We can then finish simplifying to get:

(a)

This is a function. For every input there is exactly one output. We can also check this by doing the vertical line test.

(b)

This is again a function. Even though this line is extremely steep, it still obeys the rule for a relation to be a function.

(c)

This is not a function. Using the vertical line test, it fails this when is near 0. Remember that even if it only for a few cases, if a relation fails this test even once, it cannot be a function.

(d)

While this looks like it fails the vertical line test at the peaks, there are in fact open and closed circles indicated on the graph. This means that at that particular input where there may possibly be two outputs, this graph is specially made such that it only exists where the circle is coloured in, and not where it is open. As a result, this will still be a function

For , we are in the domain , since it is inclusive of . That means:

Next, we are in the domain when . In this case, the function is , so there is nothing to substitute into. This means:

Finally, when , we are in the final domain, .

First, finding and , we simply substitute and .

To get , we simply add these equations.

Now, we calculate .

Expanding and simplifying gives:

As shown, , as there is an extra term in .

Firstly, for , we get:

When a negative number is raised to an even power, it becomes positive, whereas when it is raised to an odd power, it stays negative. Keeping this fact in mind, the expression simplified to this:

By factorizing the negative from the denominator and placing it out the front we can get:

We can now note that this is the same as the original function with a negative out the front. This hence tells us that:

Let’s start with the value . Substituting this into the equation gives us this:

The equation gets reduced to a quadratic. Solving this gives:

Since there are two possible values of that apply to one value of , this equation is not a function. For a relation to be a function, the condition is that each value (assuming it is the independent variable) maps to only one value. Furthermore, if we look at the graph, we can easily see that it doesn’t pass the vertical line test.

Insert graph of x2y2 - 3x^2 + y^2 = 0

Remember with relations, we have inputs and outputs. The numbers in set are the possible inputs, while set contains all possible outputs.

One example of how works, is if we take the input 3. In set , we can see there are three possible numbers that have a factor of 3, those being 6, 9, and 12. Following this same logic, we can draw a mapping diagram.

Based on this, we can conclude that is not a function. This is because the inputs are mapping to many values.

To make this simpler, we should first try to make the subject of this equation.

Now, we substitute each value of , and get our results. First, notice that when we substitute or , we get an undefined answer, since we cannot take the square root of a negative number. This is fine, since if you look at the graph of this relation, there is no real value for between and .

For the next values of , we will get real values, so this carries on as normal.

An interesting note is that there is exactly one value of at the point , and two solutions for every other value of . These points of interest can be clearly seen in the graph of this equation.

Graph x^2/4 - y^2/9 = 1

First, find an expression for .

A similar expression can be made for .

Now, putting this together to form the left hand side of the equation, we can show the statement holds true.

Substituting in all the values of gives us these answers:

We can see that as becomes larger, is not becoming much bigger. This is an indication that the function approaches a limiting value as becomes bigger. To figure this value out, let’s keeping substituting in bigger values of . Let’s try .

If we choose a bigger , we see that this answer doesn’t change. In fact, this value is a constant known as Euler’s number (not to be confused with Euler’s constant), but more simply called . This is a type of transcendental number like , and is very important in many topics of maths.