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# Finding composite functions

Through a worked example involving f(x)=√(x²-1) and g(x)=x/(1+x), learn about function composition: the process of combining two functions to create a new function. This involves replacing the input of one function with the output of another function. Also note that the composition of two functions is typically not the same as their composition in the reverse order. Created by Sal Khan.

## Want to join the conversation?

• Would one be able to define the new function as its own function.
For example, the new f(g(x)) would simply be h(x) instead?
• There is an addition sign separating the two terms, so cancelling them would not be possible
• I am lost, bad at math and give up oof oof ooof oooof I try but I suck :((((
• keep going! I feel the same way sometimes, a snack usually helps
• How do you find a composite function when one is missing for example if the question is "if g(x) = 2x and f(g(x))=3/x, what is f(x)?"
• couldn't sal simplify the answer in ?
• Hi Mirghani,

You can simplify the expression, but it will still be a hairy expression.

Let's see how it turns out-
f(g(x)) = sqrt (x/x+1)^2 - 1
= sqrt x^2/(x+1)^2 - 1
= sqrt x^2/(x+1)^2 - (x+1)^2/ (x+1)^2
= sqrt x^2 - (x+1)^2/ (x+1)^2
= sqrt x^2 - (x^2 +2x + 1)/ x^2 +2x + 1
= sqrt x^2 - x^2 -2x - 1 / x^2 +2x + 1
= sqrt -2x -1/ x^2 + 2x +1

You can see that it is still is a hairy number and will not help us much if we factored and simplified it.

I hope that this helps.

Aiena.
• At around the mark, isn't your answer not simplified enough? The three basic rules of simplifying a radical state that the radicand must not have any perfect square factors, there can't be a radical in the denominator of a fraction, and the radicand can't be a fraction. So, to simplify Sal's answer further we would do sqrt((x^2-1)/(1+sqrt(x^2-1)))*(((1-sqrt(x^2-1))/(1-sqrt(x^2-1))). Therefore, we would get (x^2-1-sqrt(x^2-1))/x^2.

• I would agree with you...a simplified answer would have no radical in the denominator. Though, when I did it I got: (x^2 +1-sqrt(x^2-1))/x^2.
Sal doesn't consistently simplify the radicals in his videos, which is frustrating.
• For f( g(x)), couldn't we just plug in the value of g(x)?
• You could if you were looking at a particular value of x. In this example, Sal was simply using the two functions to build a single function. This is useful in that if you knew a value for x, you could now evaluate a single function rather than having to successively evaluate two different functions. If you need to evaluate the composition at many different input values, generating a composition function algebraically is often more efficient.
• I was able to further simplify f(g(x)) =
sqrt((x^2/(1+x)^2) + 1)
Normalize the 1 = (x+1)^2/(x+1)^2 and expand quadratics
sqrt((x^2-x^2+2x+1)/(x+1)^2)
= sqrt(2x+1) / (x +1)

Is this correct?
• Almost. In the video, it's sqrt((x^2/(1+x)^2) - 1), not sqrt((x^2/(1+x)^2) + 1). Therein lies your mistake; sqrt(-1) is an impossible number. However, all the math beyond that small error works out. (Don't worry, I fell for it too.)
• Is it possible to do this mentally? Or is better to write the whole thing out? What do you guys think?
• With practice, you will most likely be able to find composite functions mentally. This may not happen for all problems, but for some, it certainly will.

For example, if f(x) = x + 1, and g(x) = x^2, finding f(g(x)) wouldn't most likely be regarded as hard, since you can simply substitute the x^2 in to get f(g(x)) = x^2 + 1

However, if you were given a harder example, such as f(x) = (x + tanxsecx - x!/sqrt(x)) and g(x) = cscx * arccos(x), then finding the composite function mentally would probably be harder.
• For the second example is it mathematically wrong to say f(x)/1+f(x)= 1/2? Since f(x)/f(x) will = 1 since by definition each function input can only have one output, and you're adding 1 to the denominator? Or is my logic flawed?
• If you are asking can you change f(x)/[1+f(x)] into f(x)/1 + f(x), the answer is no.

Why?
1) The denominator has terms. We can't split the fraction up by the individual terms in the denominator. If it had been multiplication, then you could.

2) f(x)/1 + f(x) = 2*f(x), so it is not equal to the original problem.

3) If you are asking can you divide out the f(x)'s in f(x)/[1+f(x)] to create 1/(1+1), the answer is no. When we reduce fractions, we divide out common factors (items being multiplied). The denominator here has terms. We can't divide out individual terms.

Hope this helps.