Difference quotient for quadratic function

[zolton5971]

Find the difference quotient f(x+h)-f(x)/h
Where h\ne 0, for the function below

f(x)=5x^2+4

Simplify your answer as much as possible.

How do I do this?

 

[MarkFL]

First, what is:

\(\displaystyle f(x+h)\) ?

Use the same technique from your previous problem to find a given function with a new input. :D

 

[zolton5971]

Is it f(x+0)-f(x)/0?

 

[MarkFL]

Is it f(x+0)-f(x)/0?

No, that would only be true for $h=0$, but you were told that $h\ne0$.

What you need to do is find $f(x+h)$, subtract from this $f(x)$, and then divide the result by $h$. I have to run for a few hours, so if anyone else wants to help with further questions in this thread, please feel free to do so.

 

[zolton5971]

I can't seem to figure it out!

 

[MarkFL]

I can't seem to figure it out!

Please post your attempt at finding $f(x+h)$...

 

[soroban]

Hello, zolton5971!

Find the difference quotient [tex]\frac{f(x+h)-f(x)}{h}[/tex]
for [tex]f(x)\:=\:5x^2+4[/tex]

There are three steps to the Difference Quotient.

(1) Find [tex]f(x+h).[/tex]
. . Replace [tex]x[/tex] with [tex]x+h[/tex] ... and simplify.

(2) Subtract [tex]f(x)[/tex], the original function ... and simplify.

(3) Divide by [tex]h[/tex] ... factor and reduce.Here we go!

[tex](1)\;f(x+h) \;=\;5(x+h)^2 + 4[/tex]
. . . . . . . . . . [tex]=\; 5(x^2+2xh + h^2) + 4[/tex]
. . . . . . . . . . [tex]=\;5x^2 + 10xh + 5h^2 + 4[/tex]

[tex](2)\;f(x+h)-f(x) \;=\;(5x^2+ 10xh + 5h^2 + 4) - (5x^2 + 4)[/tex]
. . . . . . . . . . . . . . . .[tex]=\;5x^2 + 10xh + 5h^2+ 4 - 5x^2 - 4[/tex]
. . . . . . . . . . . . . . . .[tex]=\; 10xh + 5h^2[/tex]

[tex](3)\;\frac{f(x+h)-f(x)}{h} \;=\; \frac{10xh +5h^2}{h}[/tex]
. . . . . . . . . . . . . . . . .[tex]=\;\frac{5h(2x+h)}{h}[/tex]
. . . . . . . . . . . . . . . . .[tex]=\; 5(2x+h)[/tex]
There!