## Evaluating Functions

Given g of x is equal to
8x plus 2, find g of zero, g of negative 5,
and g 2x plus 3. So all this is, when they say
g of x is equal to 8x plus 2, this is a function definition. They’re saying
you give me any x, and you input it
into my function g, and my function g will output,
will associate with that number x, another number that
happens to be 8x plus 2. This is one way to
create an association. It’s going to associate
if you were to put 1 here. If this x is 1, you’re going
to do 8 times 1 plus 2. So you would
associate 1 through g. You’d associate 1 with 8
times 1 plus 2, which is 10. So it’s just another way
of creating association that you could do with
sets of ordered pairs or you could do it graphically. But this is one of
the most typical ways you’ll see a function defined. It says, look,
input an x over here and I’m going to
output 8x plus 2. So with that out of
the way, let’s think about what the g of 0 is. g of 0 literally
just means we’re inputting 0 into this function. And this function says,
whatever your inputting, I’m going to output
8 times that plus 2. So g of 0 is going to be–
let me write it this way so it becomes clear
what I’m doing. g of 0 is going to be 8
times is 0 plus 2. So wherever we saw the
x, we replaced it what it with whatever we’re inputting. That is our x in this
situation right over here. And so this is equal
to 8 times the 0 plus 2 is just equal to 2. So g of 0 is equal to 2. Let me just write that
here. g of 0 is equal to 2. Now let’s try g of negative 5. So let me write it this way. I’ll write it, g of
negative 5 is equal to– I’ll try to do the
color-coding a little bit better this time. It’s going to be equal
to 8 times our x, which in this case, is negative 5. 8 times negative 5 plus 2. We’re inputting x– We’re
inputting a negative 5. And we’re going to output
8 times negative 5 plus 2. Wherever we saw the x, we just
replace it with a negative 5. And this is just
equal to 8 times negative 5 is negative 40 plus
2 is equal to negative 38. So I’ll write this
over here on the right. We have g of negative 5
is equal to negative 38. Now, this last one is a
little bit interesting. So we just did this one. Now, let’s do g of 2x plus 3. So now they’re not
inputting a number. They’re inputting an expression. Now, what we have to do is
everywhere where we see x here, we’re going to replace
it with that expression. If I did g of a circle–
let me write this down. If I did g of an orange
circle right over here. By this function
definition, if I’m in putting an orange
circle, I should output an orange circle 8
times the orange circle, which doesn’t make
a lot of sense. But hopefully you get the idea
that whatever you’re inputting, you’re just going to multiply
it times 8 and add two. So if we input 2x plus 3, we’re
going to output this times 8 plus 2. Wherever we see an x, we replace
it with this right over here. So let’s try that out. So now we have g of 2x plus 3
is going to be equal to 8 times whatever we’re
inputting, eight times whatever inputting plus 2. If this was an orange,
we’d put an orange here. If this is a negative 5,
we put a negative 5 here. If this is a 0, we
put a zero here. In this case, we’re inputting
2x plus 3 into our function. So we put 2x plus 3 where
x is right over there. So 2x plus 3. Whatever we input, this function
says multiply by 8 and add 2. We’re inputting 2x plus
3, multiply it by 8, and then add 2. And so this is going to be equal
to– we just distribute the 8. And you get 16x plus 24. And then you still have
this plus 2 out over here So this is going to be
equal to 16x plus these two. You add them up, you get 26. So we can write
over here, and I’ll do that same color so
it can be consistent. We have g of 2x plus 3
is equal to 16x plus 26. And we’re done. I know it’s going
to be confusing because you’re inputting–
you’re replacing x with something
that has x in it, but you really
should just view this as if this was an orange ball
or a star or even if this had y’s is it. Wherever you see an
x, you replace it with whatever you
are inputting it. In every instance, we
replace the x with 2x plus 3.