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# 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) ?(37 votes)
- 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.(100 votes)

- At1:09Sal 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?(10 votes)
- 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(30 votes)

- At2:28why is it that 7x * x^3 equals to x^4 ?(3 votes)
- 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)(0 votes)
- 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.

(a+b)(c+d) = ac+ad+bc+bd(23 votes)

- what do you mean by (fx) =7x -5 is this a formula(4 votes)
- This function f(
`x`

) =7`x`

−5 means that each time we plug in a value of x we would multiply it by 7 then subtract 5.

f(`x`

)=7`x`

−5 → This our function with just x.

Lets try substituting different values for x

f(`3`

)=7`3`

−5=21−5=16 → If substitute x by 3, this is what we get.

f(`8`

)=7`8`

−5=56−5=51 → If substitute x by 8, this is what we get.(7 votes)

- In2:14

Why x^3 * 7x = 7x^4 !

Why there's fourth power. I can't get it :((3 votes)- We have x³·7x. That is, x·x·x·7·x. We can rearrange it into 7·x·x·x·x, which is 7x⁴.(4 votes)

- 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?(4 votes)
- 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(2 votes)

- 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?(3 votes)
- What happens when we have a^2+a^3? Can we simplify that?(3 votes)

- 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!!(7 votes)

- 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

where`\forall functions f and x \in D(f) and y \in R(f):`

y=f(x) <=> x=(f^-1)(y)`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?(3 votes)- 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).)(3 votes)

## Video transcript

f of x is equal to 7x minus 5. g of x is equal to x to
the third power plus 4x. And then they ask us to find f
times g of x So the first thing to realize is that this
notation f times g of x is just referring
to a function that is a product of f
of x and g of x. So by definition, this notation
just means f of x times g of x. And then we just
have to substitute f of x with this definition,
g of x with this definition, and then multiply out these
algebraic expressions. f of x is right over there. And g of x, is right over there. So let's do it. So this is going to be
equal to-- switch back to the orange color. It's going to be
equal to f of x, which is 7x minus 5
times g of x, and g of x is x to the
third power plus 4x. And you could-- we're
multiplying two expressions that each have two terms. You could use FOIL if you like. I don't like using
FOIL because you might forget what
it's even about. Foil is really just using the
distributive property twice. So for example, you
take this expression. Whatever you have out here,
if you had a 9 out here, or an a, or an x, or anything. Now you have 7x minus 5. If you're multiplying it
times this expression, you would multiply this
times each term over here. So when you multiply 7x
minus 5 times x to the third, you get-- I'll
write it this way. You get x to the third times--
actually, let me write it the other way. You get 7x minus 5
times x to the third. And then you have plus
7x minus 5 times 4x. And now we can do the
distributive property again. We're not normally
used to seeing the things we distribute
on the right hand side. It's the same exact idea. We could put the x to
the third here as well. And when we distribute,
you multiply x to the third times 7x
and times negative 5. x to the third times 7x
is 7x to the fourth power. X to the third times negative
5 is minus 5x to the third. And then you do it over here. You distribute the
4x over the 7x. 4x times 7x is plus 28x squared. 4x times negative
5 is minus 20x. And let's see if we
can simplify this. We only have one fourth degree
term, one third degree term, one second degree term,
and one first degree term. Actually, we can't
simplify this anymore. And we're done. This is the product of those
two function definitions. This is f times g of x. It is a new function created
by multiplying the other two functions.