>> Welcome to the Cypress College Math Review
on Difference Quotients.
Here's the graph of a function, y = f(x).
A secant line is a line through two points on a curve.
Here's our two points.
Delta x is a change in x. So we have our first point
with coordinates x, f(x) and our second point
with coordinates x + delta x and f
of the quantity x + delta x. We want to find the slope
of the line through those two points.
Here's the formula for slope.
Now let's apply it.
So the slope will be rise over run, so the change
in the y coordinates over the change in the x coordinates,
which gives us the formula for the difference quotient.
So the difference quotient is the slope of the secant line.
A lot of instructors, instead of using delta x, use h,
and that's what we'll use in the problems.
So instead of using delta x, we will use h,
so this is the formula that we will use
for the difference quotients in the problems.
Let's find the difference quotient for this function.
Here's our formula we're going to need to use.
Let's first find f of the quantity x + h. So we're going
to replace the variable with x + h. We distribute
and we get 5 x + 5 h. Now we want our difference quotient.
So we have f of the quantity x + h
minus f(x) all over h. This is a really good idea.
Brackets minus brackets all over h. What do you put
in the first brackets?
Well, f of the quantity x + h,
so that would be 5 x + 5 h - 7 in this case.
What goes in the second brackets?
The original function.
Now get rid of the brackets.
So 5 x + 5 h - 7, distribute the negative,
so we have negative 5 x + 7 all over h,
then we have positive 5 x - 5 x, negative 7 and positive 7,
so we get 5 h over h. The h's cancel,
so we get 5 with the restriction,
of course, that h cannot be 0.
Let's find the difference quotient
for this quadratic function.
Every time you see an x, you're going to replace it with x + h.
And now we simplify the right-hand side.
Watch out, exponents before multiplication,
so we need to square out this binomial.
You cannot distribute the 3.
Exponents come before multiplication.
So we square out the binomial.
What do you get when you square out a binomial?
When you square out a + b, you get what?
A squared + 2 a b + b squared, right?
So in this case, that would be x squared + 2 x h or h x,
either one, + h squared.
Now I distribute the negative 2, so negative 2 x - 2 h + 3.
Now we distribute the 3.
And any time you change what you're doing,
you need to change your label on the left-hand side.
Now we're doing something else.
We have to change our label.
Now we're doing the whole difference quotient,
so we have to change our label on the left-hand side.
So we wrote down our label.
Our new label is the whole formula
for the difference quotient.
Brackets minus brackets.
So we have the brackets minus brackets all
over h. What do we put in the first brackets?
f of the quantity x + h, what we were just working on.
What do we put in the second brackets?
The original function.
Now we get rid of the brackets.
So we distribute the negative.
So we have a positive 3 x squared
and a negative 3 x squared.
We have a negative 2 x
and a positive 2 x. We have a positive 3 and a negative 3.
Now we can factor out an h from every term that's left
and we're left with 6 x + 3 h - 2 all over h. The h's cancel,
so we're left with 6 x + 3 h
minus 2 with the restriction that h cannot be 0.
So in this case, f
of the quantity x + h really can't be simplified.
So here we go with the difference quotient.
So brackets minus brackets, and in this case,
the brackets don't serve any purpose.
All right, there's two different ways
to simplify this complex fraction, and I'm going
to do the problem twice so you can see the two different ways.
The first way is to multiply it by the number 1
which eliminates the complex fraction.
So I determine the LCD of these fractions,
and the least common denominator would be x times the quantity x
+ h. So I multiply the numerator
and the denominator by that expression.
Of course, you can only multiply an expression by the number one,
and that's equal to number one, distribute this
to the first term in the numerator, which gives me 3 x,
distribute it to the second term in the numerator
and it gives me 3 times the quantity x + h.
In the denominator, it's h times x times x + h,
and it's all factored and you want it factored,
so we leave it like that.
In the numerator, we distribute the negative 3, 3 x - 3 x,
so we have negative 3 h over h x times x + h. The h's cancel,
so we end up with negative 3
over x times the quantity x + h with, again,
the restriction that h cannot be 0.
Referring back to the same problem,
to add or subtract fractions,
where the least common denominator is simply the
product of the denominators, you just cross-multiply, yes,
a times d + b times c over b times d, yes,
to add fractions, okay?
So we'll do that here.
3 times x minus, in this case, 3 times the quantity x + h all
over x times the quantity x + h, yes.
Now, I'm dividing by h. I really don't want to do this all
over h. That gives, again, a complex fraction.
So to divide by h, that's the same thing
as multiply by 1 over h, yes?
Well, why write it like that?
Multiply it by 1 over h. Why don't I simply write it
as h times it, yeah?
Multiply h times the denominator.
Good. And then we do the same thing we did earlier,
since we've already done this problem, and we get negative 3 h
over h x times the quantity x + h. The h's cancel
and we get negative 3 over x times the quantity x + h
with the restriction that h cannot be 0.
Here are some problems to try on your own.
Pause the video, try these problems, then restart the video
to check your answers.
Here are the answers to the problems.