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## Algebra 1

### Course: Algebra 1 > Unit 8

Lesson 1: Evaluating functions- What is a function?
- Worked example: Evaluating functions from equation
- Evaluate functions
- Worked example: Evaluating functions from graph
- Evaluating discrete functions
- Evaluate functions from their graph
- Worked example: evaluating expressions with function notation
- Evaluate function expressions

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# Worked example: evaluating expressions with function notation

Evaluating -2⋅f(-6)+g(1) given the graphs of f and g. Created by Sal Khan.

## Want to join the conversation?

- How can we apply this information in real life? I mean, where do we see these types of graphs in our day-to-day lives?

I don't get how this would be useful.(41 votes)- We Don't Learn Only Things That We can use in life but we also learn things to enrich our minds so that we can stretch them and learn how to problem solve and become independent instead of dependent! Do You know what I Mean?(16 votes)

- is there a reason that sal always uses the x as the input and receives a y value in return and not the other way around(8 votes)
- Tradition and the Cartesian Plane which uses x as the input and y as the output(15 votes)

- I don't know where is the best place to ask this question , and I am sorry for any bad English.

Can i have a function inside it self, is there a way to solve this or am I making this up ?

for example f(x)=f(x)+2+x; if I replace it with f(2) would there be a way to solve this ?(2 votes)- Your example doesn't really involve a function. Notice that we can subtract f(x) from both sides to get 0=2+x.

Now subtract 2 from both sides to get x= -2. So we just have an equation of one variable that we can solve.

What we*can*do is define functions recursively. For example, we can define a function over the whole numbers:

f(0)=1

f(n)=n*f(n-1)

So if we want to find f(3), this is 3•f(2)=3•(2•f(1))=3•2•(1•f(0))=3•2•1•1=6.

If you meant something else, comment back.(15 votes)

- Around1:16wouldn't it be -14+-5=-9 instead of -14+-5=-19?(6 votes)
- Same signs add numbers and keep sign, so since both are negative, add numbers (14+5) and keep the negative sign to get -19.(5 votes)

- The questions in the video are so easy. Can they be harder?(5 votes)
- how old do you have to be to do this stuff?(0 votes)
- There's no age limit! However, I think most schools teach this in grades 8-9.(12 votes)

- How do you tell the difference between function notation and regular multiplication?

Ex: f(x) is a function or f*x? Like in an actual algebraic complex equation?(2 votes)- Once functions have been introduced, you should always assume f(x) is referring to a function named "f" with input value of "x". If it was multiplication, it would be written as "fx".(5 votes)

- Why can I never remember this?(4 votes)
- idk how to do this fr 💀(1 vote)
- What is happening here is basically you substituting the variable function [ex: f(x)] with the graphs. For the example -2⋅f(-6)+g(1), you find where -6 and 1 are on the graphs, since those are the x values, then substitute the values in the equation. Since g(1) lines up with -5, that is the value. After you substitute in the values just solve as a normal equation and you have the answer. Hope this cleared things up.(4 votes)

- at1:06, why does Sal say g of 1 = -5, when there is also 7? Which one is it? Thanks(2 votes)
- No, there isn't also a 7.

g(1) means you need to find the output value of g(x) when x=1. g(x) is the red dashed line. When x=1, the y=value that is on the red dashed line is -5.

I don't see where you get the 7. The other line is f(x) and f(1)=8, not 7.(1 vote)

## Video transcript

We're asked to evaluate
negative 2 times f of negative 6 plus g of 1. And they've defined,
at least graphically, f of x and g of x here below. So let's see how we
can evaluate this. Well, to do this, we
first have to figure out what f of negative 6 is. So our input into our
function is negative 6. And we'll assume that's
along the horizontal axis. So our input is negative 6. And based on our function
definition, f of negative 6 is 7. So this thing. Let me write this down. f
of negative 6 is equal to 7. And what is g of 1? Well, once again,
here's our input axis. And then the function
says that g of 1, which is right over
there, is negative 5. g of 1 is equal to negative 5. So this statement simplifies
to negative 2 times f of negative 6, which is 7. So times 7 plus g of
1, which is negative 5. So plus negative 5, which
simplifies to-- let's see. Negative 2 times 7 is
negative 14 plus negative 5, which is negative 19. And we are done.