Last time we saw that we can replace y in an equation with f(x) when y is alone on the left side of an equation:
y = f(x) = 3x² – 5x + 1
The above is an example of the function definition. Once defined, you replace all the x‘s on the right side with whatever is in the brackets on the left side, even if it is not a number. For example,
f(2) = 3(2)² – 5(2) + 1 = 3
f(a) = 3a² – 5a + 1
Even if the thing in the brackets is another expression, for example, an expression that is used in calculus a lot is x + h:
f(x+h) = 3(x + h)² – 5(x + h) + 1
And you can even use another function of x inside the brackets of another function. Like x, the letter f is used in the first instance for a function, but if other functions need to be defined as well, other letters are used:
f(x) = 3x² – 5x + 1
g(x) = x² – 7
f[g(x)] = f(x² – 7) = 3(x² – 7)² – 5(x² – 7) + 1
g[f(x)] = g(3x² – 5x + 1) = (3x² – 5x + 1)² – 7
The domain of a function is all the valid values of x that can be used. Many times, the domain of a function (like f(x) and g(x) above) is just any real number. But there are functions where you cannot use just any number. For example, consider
\[{f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{{x}{-}{2}}
\]
There is one value of x you cannot use. That value is 2 because that will make the denominator 0, and as you know, this will bring the maths police to your door. So the domain of this function is all real numbers except for 0.
Now consider
\[{f}{(}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{x}{-}{2}}
\]
Another illegal operation is taking the square root of a negative number. The requirement for this function is that x – 2 has to be 0 or greater. For this to be true, x must be greater than or equal to 2. The phrase ” greater than or equal to” can be replaced by the maths symbol ≥. So the domain of this function is x ≥ 2.
There are other reasons why the domain of a function is restricted, but the most common things to look for is dividing by 0 or taking the square root (or any even root) of a negative number.