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Course: Algebra (all content) > Unit 7
Lesson 15: Composing functions (Algebra 2 level)- Intro to composing functions
- Intro to composing functions
- Composing functions
- Evaluating composite functions
- Evaluate composite functions
- Evaluating composite functions: using tables
- Evaluating composite functions: using graphs
- Evaluate composite functions: graphs & tables
- Finding composite functions
- Find composite functions
- Evaluating composite functions (advanced)
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Intro to composing functions
This video is about composing functions, which is the process of building up a function by composing it from other functions. It explains how to evaluate the composition of functions step by step, using examples with three different function definitions: f (x), g(t), and h(x). Created by Sal Khan.
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could you still answer f(g(x)) without knowing what "x" is?(17 votes)
Yes. You can literally plug in the whole equation for g(x) in for f(x). For example:
f(x) = 2x + 1 and g(x) = 4/x
Then to solve for f(g(x)), you would plug in g(x) (the whole formula), in to f(x) for x. So...
f(g(x)) = 2(4/x) + 1 = 8/x + 1
This is just a simple example but you can do it with many more complicated formulas as well.
For example:
(given f(x) = 2x+3 and g(x) = -x^2 + 5
( f o g)(x) = f (g(x))
= f (–x^2 + 5)
= 2(g(x)) + 3
= 2(–x^2 + 5) + 3
= –2x^2 + 10 + 3
= –2x^2 + 13
Yes I know that these formulas are not as complicated as many people doing Algebra 2 could do, but I was just giving another example...
Hope this helps... :)(74 votes)
can a function have more than one variable? like: f(a,b) or something(27 votes)
Absolutely! A function with more than one variable would be called a multivariable function. and that is actually the correct way of notating it! A function with 2 variables would make a 3D object, 3 would make a 4D object, etc. Great question!(38 votes)
will i need this for real life. plzzz help i need to kno(0 votes)
Yes; say that I sell cars. In the function f(x) = y, the input (or x) is the number of cars being made in the year and the output (or y) is the price one needs to sell a car for. And then I will need to calculate the least amount of cars I would have to sell in order to earn enough money to pay my employees. The price of the car in that year will be P and the number of cars I will need to sell will be S, so I will now have a new function called g(P) = S.
However, it's too much unnecessary work for me to have those two separated because P, the price one needs to sell a car for, equals my definition of y, which equals f(x). So instead of going through two functions, I can now use g(f(x)) to find out how many cars I need to sell in the least in each year.(63 votes)
Hey guys
Is it necessary to complete Algebra 2 before staring Precalculus ?(3 votes)
Yes. There are foundation skills in Algebra 2 that you need to know in precalculus.(10 votes)
At2:24where did you get (-3)^2-1 from?(6 votes)
Sal is plugging in (-3) for x in f(x). f(x) = x^2 - 1, so plugging in x makes f(x) = (-3)^2 - 1. He also gets the -3 from g(2) because g(2) is the input into the function f(x).(6 votes)
- WHAT IS THIS, I'm totally CONFUSED-(8 votes)
is the chart and graph the same for each example?(7 votes)- I have a question that is more general. I'm taking Precalculus next year, and was planning on studying in advance through Khan Academy. Has anyone taken Precalculus before? If so, does the curriculum in Khan Academy align with what is being taught at school?(5 votes)
- Precalc Honors for me included topics such as:
Graphing linear, exponential, quadratics, cubics, quartics, stuff with higher degrees, logs, natural logs, powers of e.
Solving inequalities, factoring, difference of squares, difference quotient, stretch/shrink of graphs, vertical/horizontal shifts of graphs, reflections of graphs, horizontal/vertical asymptotes, properties of logs, solving logs, unit circle, radians, degrees, linear velocity, angular velocity, graphing trig functions, angle of depression/elevation, domain/range of trig functions, solving triangles, rule of sines, rule of cosines, equations/graphing/rules of: roses, limacons, cardioids, parabolas, ellipses, hyperbolas, and parametrics. Limits, sigma notation, geometric and arithmetic, half life, apert formula, a=pe(1+r/n)^nt thing, explicit/recursive.
This isn't a comprehensive list, since I don't recall everything, and also what I learn may not be what you learn, but this should give you a general idea.(3 votes)
I have a few questions, why is it called class 12 math (India)? Is it like Indian math or something? Also I am in 8th grade, do you think I am too young to be learning this? Is this only for 12th graders?(4 votes)- KhanAcademy has multiple courses. They customize them to a target audience. However, they don't create new videos for each course. A course for the India curriculum likely uses this video as does the Pre-Calculus course for the US.(3 votes)
(Ignore top please)
On my study guide, it says that F(x)= 8 =x(x-5)(2x-4). I know that an easy guess for x is 1, and if you write it out you can confirm it without strenuous arithmetic. but I don't understand how it goes from F(x) is 8, and 8=F(1), to saying that (x-1) is a factor of the equation 0 = 2x^3 - 14x^2 + 20x - 8
(2x^3 - 14x^2 + 20x is the standard form of x(x-5)(2x-4))
....Oh wait, does any real possible answer to x minus itself make a zero? And if so,
Also, is there an easier way to find a possible x (such as the 1 in the above equations) than guessing?(4 votes)
ƒ(x) = x(x - 5)(2x - 4)
ƒ(1) = 8
That means when you plug in 1 for "x" in the above expression, you will get 8.
Remainder Theorem tells us that when we divide ƒ(x) by a linear binomial of the form (x - a) then the remainder is ƒ(a). We know that ƒ(1) = 8. It follows that if we divide ƒ(x) by (x - 1), then our remainder is 8. We can make (x - 1) a factor of ƒ(x) if we add something to the function that will get rid of the remainder. Since the remainder is 8 and we want to get rid of that, we subtract 8 to get:
ƒ₁(x) = 2x³ - 14x² + 20x - 8
If we divide by (x - 1) our remainder is:
ƒ₁(1)
Which we note is 0, because the first 3 terms are from the original function ƒ(x) and that already yielded 8, and when we combine that with the remaining -8, we get 0. Therefore (x - 1) is indeed a factor of 2x³ - 14x² + 20x - 8.
Your second question asks if there is an easier way to solve the following equation:
x(x - 5)(2x - 4) = 8
Here is the systematic algebraic way to do it:
First expand the LHS:
2x³ - 14x² + 20x = 8
Subtract 8:
2x³ - 14x² + 20x - 8 = 0
We solve this equation by factoring it. From our analysis above, we know that (x - 1) is a factor of the polynomial, so we want to divide the polynomial by (x - 1) and find the quotient. We can do this with synthetic division. This yields:
(x - 1)(2x² - 12x + 8) = 0
We can continue to search for roots by finding the roots of the quadratic:
2(x - 1)(x² - 6x + 4) = 0
(x - 1)(x² - 6x + 4) = 0
The quadratic is not factorable. The quadratic formula yields roots 3 ± √5.
Therefore the solutions to the equation:
x(x - 5)(2x - 4) = 8
are:
x ∈ {1, 3 ± √5}
Comment if you have questions.(4 votes)
2:24Video transcript
Voiceover:So we have three different function definitions here. This is F of X in blue, here
we map between different values of T and what G of T would be. So you could use this as
a definition of G of T. And here we map from X to H of X. So for example, when X is equal to three, H of X is equal to zero. When X is equal to one,
H of X is equal to two. And actually let me number this one, two, three, just like that. Now what I want to do in
this video is introduce you to the idea of composing functions. Now what does it mean
to compose functions? Well that means to build
up a function by composing one function of other functions or I guess you could think of nesting them. What do I mean by that? Well, let's think about
what it means to evaluate F of, not X, but we're
going to evaluate F of, actually let's just start
with a little warm-up. Let's evaluate F of G of two. Now what do you think this is going to be and I encourage you to pause this video and think about it on your own. Well it seems kind of daunting at first, if you're not very
familiar with the notation, but we just have to
remember what a function is. A function is just a mapping from one set of numbers to another. So for example, when
we're saying G of two, that means take the number two, input it into the function G and
then you're going to get an output which we are
going to call G of two. Now we're going to use
that output, G of two, and then input it into the function F. So we're going to input
it into the function F, and what we're going to get is F of the thing that we
inputted, F of G of two. So let's just take it step by step. What is G of two? Well when T is equal to two,
G of two is negative three. So when I put negative three
into F, what am I going to get? Well, I'm going to get negative
three squared minus one, which is nine minus one which
is going to be equal to eight. So this right over here is equal to eight. F of G of two is going
to be equal to eight. Now, what would, using
this same exact logic, what would F of H of two be? And once again, I encourage
you to pause the video and think about it on your own. Well let's think about
it this way, instead of doing it using this little
diagram, here everywhere you see the input is X,
whatever the input is you square it and minus one. Here the input is H of
two, and so we're going to take the input, which is H
of two, and we're going to square it and we're going to subtract one. So F of H of two is H of
two squared minus one. Now what is H of two? When X is equal to two, H of two is one. So H of two is one, so since
H of two is equal to one, this simplifies two one squared minus one, well that's just going to be one minus one which is equal to zero. We could have done it
with the diagram way, we could have said, hey
we're going to input two into H, if you input
two into H you get one, so that is H of two right over here. So that is H of two, and then
we're going to input that into F, which is going
to give us F of one. F of one is one squared
minus one, which is zero. So this right over here is F of H of two. H of two is the input
into F, so the output is going to be F of our
input, F of H of two. Now we can go even further,
let's do a composite. Let's compose three of
these functions together. So let's take, and I'm doing
this on the fly a little bit, so I hope it's a good
result, G of F of two, and let me just think
about this for one second. So that's going to be G of F of two, and let's take H of G of
F of two, just for fun. Now we're really doing
a triple composition. So there's a bunch of
ways we could do this. One way is to just try to
evaluate what is F of two. Well F of two is going to be
equal to two squared minus one. It's going to be four minus one or three. So this is going to be equal to three. Now what is G of three? G of three is when T is equal
to three, G of three is four. So G of three, this whole thing, is four. F of two is three, three of G is four. What is H of four? Well we can just look back
to our original graph here. When X is four, H of four is negative one. So H of G of F of two, is
just equal to negative one. So hopefully this you
somewhat familiar with how to evaluate the composition of functions.