Question

Difference quotient for f(x) = x^2+x

Answer 1

Hey Alex!
Do you remember your slope formula?\[\large\rm m=\frac{y_2-y_1}{x_2-x_1}\]Well that's all our difference quotient really is,
but it's written with function notation so it looks a little fancier.

Answer 2

\(\large\rm y=x^2+x\)

So with the old notation,
we might label our y2 and y1 like this,
\(\large\rm y_1=(x_1)^2+x_1\)
\(\large\rm y_2=(x_2)^2+x_2\)

With function notation we would write the first point like this,\[\large\rm f(x_1)=(x_1)^2+x_1\]and instead of calling the second point x2,
it's going to be x1 plus some unknown distance to the right.
So our x2 is actually located at x1+h.

So our second point will look like this,\[\large\rm f(x_1+h)=(x_1+h)^2+x_1\]

Answer 3

So then subtracting our y's: \(\large\rm f(x_1+h)-f(x)\)
and subtracting the x's below: \(\large\rm x_1+h-x_1\) (Which simplifies to h),

gives us our difference quotient.

Answer 4

Sorry maybe that's more information than you needed :)\[\large\rm \frac{f(x+h)-f(x)}{h}\]

Answer 5

\[\large\rm f(\color{orangered}{x})=(\color{orangered}{x})^2+\color{orangered}{x}\]If we evaluate this function at x+h instead of x,\[\large\rm f(\color{orangered}{x+h})=(\color{orangered}{x+h})^2+\color{orangered}{x+h}\]We can see the things that we'll be plugging into the numerator.

Answer 6

Okay I'm with you so far as to plugging the x+h in to get f(x+h) = (x+h)^2 + x+h but after that I'm a little lost.

Answer 7

So we've determined that,
\(\large\rm \color{indianred}{f(x)=x^2+x}\)
and
\(\large\rm \color{royalblue}{f(x+h)=(x+h)^2+(x+h)}\)

So we're going to make the replacement in our difference quotient,\[\large\rm \frac{\color{royalblue}{f(x+h)}-\color{orangered}{f(x)}}{h}\quad=\quad \frac{\color{royalblue}{(x+h)^2+(x+h)}-\color{orangered}{(x^2+x)}}{h}\]

Answer 8

Do you remember how to expand out a square binomial?
\(\large\rm (x+h)^2=(x+h)(x+h)=?\)

Answer 9

I think so, that would be x^2 + 2xh + h^2?

Answer 10

\[\large\rm \frac{(x+h)^2+(x+h)-(x^2+x)}{h}\]Ok good job, we've expanded the square,\[\large\rm \frac{x^2+2xh+h^2+(x+h)-(x^2+x)}{h}\]Let's drop the brackets on the middle term, those aren't necessary.
And let's distribute the negative to each of the last two terms,\[\large\rm \frac{x^2+2xh+h^2+x+h-x^2-x}{h}\]

Answer 11

From here, combine like-terms.

Answer 12

If you do it correctly,
you'll be left with ONLY h stuff in the numerator.

Answer 13

Ok one sec

Answer 14

h^2+2xh+h / h?

Answer 15

Good good good.
\[\large\rm \frac{h^2+2xh+h}{h}\]From here, you can either directly divide h out of everything, or,
if you'd like you can factor an h out of each term in the numerator before doing the division.

Answer 16

Would I get h + 2x?

Answer 17

If we factor an h out of each term in the numerator, we have,\[\large\rm \frac{h(h+2x+1)}{h}\]Don't forget about that 1 on the end! :)
Factoring h out h doesn't leave you with nothing.

Answer 18

h out of h*

Answer 19

My bad, is it 2x + 1 then?

Answer 20

Dividing the h's,\[\large\rm \frac{\cancel h(h+2x+1)}{\cancel h}\quad=\quad h+2x+1\]Like that, ya?

Answer 21

Oh yeah you're right.

Answer 22

Yayyy team we did it!
It's not so bad, right?

Just a few tricky Algebra steps.

Answer 23

Yeah, you made it a lot easier to understand. Thank you so much!