f(x + h) -f(x)/h ......can someone please show me the steps and answer to this

**wayneo1688** · October 1st 2007, 12:10 PM

Find f(x+h) - f(x) for f(x) = x (squared) +5x - 1
h

**Soroban** · October 1st 2007, 12:46 PM

Hello, wayneo1688!

I assume you're new to the Difference Quotient.

Find $\frac{f(x+h) - f(x)}{h}$ . for . $f(x) \:= \:x^2 +5x - 1$

Don't let the Difference Quotient scare you.
. . Break it down into its three basic steps.

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

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

(3) Divide by $h$ . . . Factor and cancel (carefully!)

Here we go! . . . We have: . $f(x)\:=\:x^2 +5x - 1$

$(1)\;\;f(x+h) \;=\;(x+h)^2 + 5(x+h) - 1 \;=\;x^2 + 2xh + h^2 + 5x + 5h - 1$

$(2)\;\;f(x+h) - f(x) \;=\;(x^2 + 2xh + h^2 _ 5x + 5h - 1) - (x^2 + 5x-1) \;=\;2xh + h^2 + 5h$

$(3)\;\;\frac{f(x+h)-f(x)}{h}\;=\;\frac{2xh + h^2 + 5h}{h} \;=\;\frac{\not{h\;}(2x + h + 5)}{\not{h\:}} \;=\;{\color{blue}2x + h + 5}$

**KarlosK** · March 2nd 2010, 07:18 AM

I know this was posted a very long time ago, but stumbled across it trying to solve my calculus homework and wanted to say it is very inciteful and informative. Thanks!

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