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:

$$f(x+h)$$ ?

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 \frac{f(x+h)-f(x)}{h}
for f(x)\:=\:5x^2+4

There are three steps to the Difference Quotient.

(1) Find f(x+h).
. . Replace x with x+h ... and simplify.

(2) Subtract f(x), the original function ... and simplify.

(3) Divide by h ... factor and reduce.Here we go!

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

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

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