Question

compute the difference quotient of the given function

H(x)=2x^2+9

Answer 1

\[\frac{f(x+h)-f(x)}{h}\] lets take it slow one step at a time

Answer 2

\[f(x)=2x^2+9\]
\[f(x+h)=2(x+h)^2+9\]
\[=2(x+h)(x+h)+9\]
\[=2(x^2+2xh+h^2)+9\]
\[=2x^2+4xh+2h^2+9\]

Answer 3

therefore
\[f(x+h)-f(x)=2x^2+4xh+2h^2+9-(2x^2+9)\]
\[=2x^2+4xh+2h^2-2x^2-9\]
\[=4xh+2h^2\]

Answer 4

and finally
\[\frac{f(x+h)-f(x)}{h}=\frac{4xh+2h^2}{h}=\frac{(4x+2h)h}{h}=4x+h\]

Answer 5

if all this algebra looks daunting, don't fret in a week you will be able to do this problem in less than one second, using some trick called the "power rule"

Answer 6

@jessiedeee would you like me to explain the significance of the difference quotient to you?

Answer 7

@satellite73 whats the power rule?
and @Kainui please?

Answer 8

So you're in calculus, right? Have you talked about limits yet? The difference quotient when put together with a limit is what gives you a derivative, which I'll explain shortly.

Answer 9

yes I'm in calculus. I haven't learned about any of that yet. I had a summer project to do .

Answer 10

ok if you take the limit as \(h\to 0\) of \(4x+2h\) you get \(4x\) because , well because the \(h\) went to 0

Answer 11

next week you will be asked for the derivative of \(2x^2+9\) which is another way to say "find the difference quotient"
you will think "pull out the exponent of 2 as a multiplier, drop the power by 1, then get rid of the 9" and write \(2\times 2x=4x\)

Answer 12

general rule says the difference quotient (derviative) of \(x^n\) is \(nx^{n-1}\)

Answer 13

So in this picture we have a graph of y=x^2. So far so good, right?

Now what we wan't to do is find the slope of the graph at a point. This is one of the main goals of calculus, so don't forget it. So remember on a line you have the same slope everywhere, y=mx+b, right? m is the rise over the run, remember? But as we can see on this graph, the slope changes at any point.

\[slope=\frac{ rise }{ run }=\frac{ y _{f}-y _{i} }{ x _{f}-x _{i} }\]
If you don't understand the slope formula, stop me so I can explain it to you. Trust me, you want to know because like I said, calculus is all about slope of a line!

So first off, since slope is just rise/run, what is rise? That's just a change in y-value and run is a change in x-value, with me so far? So we pick a point, x, on the graph and some point a distance away from x, which is h away from x. So you see that we have two points x, and (x+h). Similarly, we have a y-value that's f(x) and f(x+h). These are our initial and final x and y values. So we plug them into the slope formula:

\[slope=\frac{ y _{f}-y _{i} }{ x _{f}-x _{i} }=\frac{ f(x+h)-f(x) }{ (x+h)-x }\]

Tell me if you have questions, but you should see now how this is an approximation of the slope between those two points. Next I'll show you how you're allowed to divide by 0 for the first time. =D