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## Algebra (all content)

### Course: Algebra (all content)>Unit 7

Lesson 14: Combining functions

# Multiplying functions

Sal solves the following problem: given that f(x)=7x-5 and g(x)=x^3+4x, find (f*g)(x). He explains that generally, (f*g)(x)=f(x)*g(x). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Could you simplify it further to: x(7x^3-5x^2+28x-20) ?
• You aren't making it any simpler by factoring out an x. If anything it complicates it.
But yes you could if you wanted to.
• At Sal says that he doesn't like using FOIL. FOIL is really easy and is much less confusing then the way Sal did the distributive property twice even though you get the same answer. Why does Sal not like FOIL?
• FOIL won't help you if you have to expand a product that isn't two binomials multiplied together; for example, two trinomials multiplied together. It's usually better to understand what you're doing instead of relying on mnemonics.

For example: (a + b + c) * (d + e + f)
= ad + ae + af + bd + be + bf + cd + ce + cf
• At why is it that 7x * x^3 equals to x^4 ?
• but 7x means X+X+X+X+X+X+X
so if you multiply it by an exponent shouldnt you get a different answer?
(1 vote)
• How do you find the product of this question? (x+1)(-5x+7)
• Alright, so we use something called FOIL, which stands for "firsts, outers, inners, lasts". When we multiply binomials like this, we sum the products of each of the components of FOIL.

• what do you mean by (fx) =7x -5 is this a formula
• This function f(x) =7x−5 means that each time we plug in a value of x we would multiply it by 7 then subtract 5.
f(x)=7x−5 → This our function with just x.
Lets try substituting different values for x
f(3)=73−5=21−5=16 → If substitute x by 3, this is what we get.
f(8)=78−5=56−5=51 → If substitute x by 8, this is what we get.
• In
Why x^3 * 7x = 7x^4 !
Why there's fourth power. I can't get it :(
• We have x³·7x. That is, x·x·x·7·x. We can rearrange it into 7·x·x·x·x, which is 7x⁴.
• What do you do if you're given a number? For example, you're given your two equations and it says: Find (f o g) (0)? Do you plug the zero into your equations or solve f*g then input the 0 into your output?
• You could do either way.
Personally, I prefer to solve both equations for the given number first, then combine them. But it doesn't really matter
• When we got the final function after solving the problem, Sal stated that we cannot simplify the expression because each variable was to a different degree. But, I thought one can subtract variables with exponents, so why can one not simplify the expression?
• What happens when we have a^2+a^3? Can we simplify that?
• is (f*g)(x) the same as (f o g)(x)?
(1 vote)
• no, (f o g)(x) = f(g(x))
but (f*g)(x) = f(x)*g(x)
i think thats what it is ,but i am not SURE!!
• Hi. I'm primary a programmer but I'm really interested in math and I think I found another possible interpretation of multiplying functions.
The key is in the notation f^-1 which is defined as
\forall functions f and x \in D(f) and y \in R(f):  y=f(x) <=> x=(f^-1)(y)
where D(f) is the domain and R(f) is the range of a function.

It gives a clue that there could possibly be f^2 meaning f(f(x)).
This would mean that (f \times g)(x) = f(g(x)).

NOTE: In latex \times is a vector cross product. I use to distinguish it from the ordinary multiplication. The same as cross product, this type of multiplication is also not commutative.

The \times operator treats similar to ordinary multiplication of real numbers:
\forall x \in R-{0}: x * x^-1 = 1 where 1 is the neutral element of multiplication.
\forall functions f: f \times f^-1 = n where n is the neutral function n(x)=x.

Is this used somewhere in the mathematics?
• Yes, that's all correct. f^2(x) is commonly taken to mean f(f(x)), similarly with higher exponents. Negative exponents refer to the inverse function, that is, f^(-2)(x) = f^(-1)(f^(-1)(x)).

What you called \times is called function composition, and is written (g ∘ f)(x) = g(f(x)). As you noted, it's not commutative, but it is associative. Whenever the compositions are defined, (h ∘ g) ∘ f = h ∘ (g ∘ f) = h ∘ g ∘ f.

In a way, the function iteration can be extended to fractional exponents as well. For example, the function g(x) = f^(1/2)(x) would be a function that satisfies g^2(x) = f(x).

Also, as a side note, the neutral function is more commonly called the identity function (and the neutral element 1 is called the identity element).

(The trigonometric functions break this convention: sin^2(x) is taken to mean sin(x)*sin(x). However, sin^(-1)(x) = arcsin x still refers to the inverse function. Arcsin is preferred over sin^(-1).)