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A function can be written as: $$y = f(x)$$ This means that, you take \(x\), and do something to it, represented by \(f\), which results in \(y\) Sometimes, a function is "Composed" of another function, and is written as follows: $$y = f(g(x))$$
A function is a mathematical expression which takes a value of \(x\), and performs and operation on it, called \(f\), to get \(y\). A function is read out loud as "y equals f of x". An important rule of functions is that no value of \(x\) can give multiple values of \(y\), but multiple values of \(x\) can give the same value of \(y\)
If \(f(x) = x^2 + 3x - 2\), find \(f(3)\) and \(f(5)\)
To find \(f(3)\), we take the number three, and replace every instance of \(x\) with it. So, \(x^2 + 3x - 2\) becomes $$3^2+3*3-2$$ $$= 9 +9 -2$$ $$ =16$$ Finding \(f(5)\) involves the exact same process $$f(x) = x^2 + 3x - 2$$ $$f(5) = 5^2 +3*5 -2$$ $$f(5) = 25 - 15 - 2$$ $$f(5) = 8$$
If you have two functions, \(f(x)\) and \(g(x)\), you can make what is called a composite function of the two, which is written as \(f(g(x))\) and said as "f of g of x". This can be confusing, so it is a good idea to take a look at the example below
Given \(f(x) = x^2 +2\) and \(g(x) = x-3\) find \(f(g(x))\) and \(g(f(x))\)
To find \(f(g(x))\), we first replace every \(x\) term in \(f(x)\) with \(g(x)\), and we get $$f(g(x)) = (g(x))^2+2$$ Next, we can plug in \(g(x)\) to get $$f(g(x)) = (x-3)^2 +2$$ $$f(g(x)) = x^2-6x+11$$ Next, we can apply a similar process to find \(g(f(x))\). First plug in \(f(x)\) for every x in \(g(x)\) $$g(f(x)) = f(x) - 3$$ Next, we can plug in \(f(x)\) as before $$g(f(x)) = x^2+2-3$$ $$g(f(x)) = x^2-1$$
You will see that the process from example two can also be applied to functions with multiple \(x\) terms
Given \(f(x) = \sqrt{x^2-3x}\) and \(g(x) = x+1\) find \(f(g(3))\)
There are two ways to find \(f(g(3))\), and the second is faster, so read both and choose the one you are comfortable with.
First Method:
First, we need to find \(f(g(x))\). To do this, we plug in \(g(x)\) for every \(x\) in \(f(x)\)
$$f(g(x)) = \sqrt{(g(x))^2-3(g(x))} $$
$$f(g(x)) = \sqrt{(x+1)^2-3(x+1)} $$
$$f(g(x)) = \sqrt{x^2+2x+1-3x-3} $$
$$f(g(x)) = \sqrt{x^2-x-2} $$
Now we plug in \(x = 3\)
$$f(g(3)) = \sqrt{3^2-3-2} $$
$$f(g(3)) = \sqrt{9-5}$$
$$f(g(3)) = 2$$
Second Method:
Instead of all of that, you can just find \(g(3)\), and use it as an input for \(f(x)\), which will get you the exact same answer. First, lets find \(g(3)\).
$$g(3) = 3+1 = 4$$
Next, we plug \(g(3)\) into \(f(x)\)
$$f(g(x)) = f(g(3)) = f(4) = \sqrt{4^2-3*4} = \sqrt(4) = 2$$
Conlcusion:
In this page you learned how to evaluate functions and composite functions, how to write them out, and how to read them aloud. If you read example three, you probably also learned a faster way of evaluating composite functions.