need help

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\(\displaystyle f(x + h) = 4\left( {x + h} \right)^2 + 5(x + h) - 5 = 4x^2 + 8xh + h^2 + 5x + 5h - 5\)
 
NO!

\(\displaystyle \left( {4x^2 + 8xh + h^2 + 5x + 5h - 5} \right) - \left( {4x^2 + 5x - 5} \right)\)=???
 
man sorry pka, I am just terrible at this factoring stuff, good thing this is my last assignment..... lets see my firned is working on it here with me too, he got,

8x + 4h + 5

(I have a feeling he is wrong too)
 
Hello, ryansmith069!

The most common problem is simply not understanding the Difference Quotient.

If \(\displaystyle f(x)\:=\:4x^2\,+\,5x\,-\,5\), then \(\displaystyle \frac{f(x + h) - f(x)}{h}\:= ?\)
That messy expression is actually a "recipe" . . . just follow its instructions.

. . [1] Find \(\displaystyle f(x\,+\,h)\) . . . Replace \(\displaystyle x\) with \(\displaystyle x+h\) ... and simplify.

. . [2] Subtract \(\displaystyle f(x)\) . . . Subtract the original function ... and simplify.

. . [3] Divide by \(\displaystyle h\) . . . . . Usually means "factor and cancel".


Here we go . . .

\(\displaystyle [1]\;\;f(x\,+\,h)\:=\:4(x\,+\,h)^2\,+\,5(x\,+\,h)\,-\,5\)
. . . . . . . . . . . . .\(\displaystyle =\:4(x^2\,+\,2xh\,+\,h^2)\,+\,5(x\,+\,h)\,-\,5\)
. . . . . . . . . . . . .\(\displaystyle =\:4x^2\,+\,8xh\,+\,4h^2\,+\,5x\,+\,5h\,-\,5\)

\(\displaystyle [2]\;\;f(x\,+\,h)\,-\,f(x)\:=\:(4x^2\,+\,8xh\,+\,4h^2\,+\,5x\,+\,5h\,-\,5)\,-\,(4x^2\,+\,5x\,-\,5)\)
. . . . . . . . . . . . . . . . . . .\(\displaystyle =\;4x^2\,+\,8xh\,+\,4h^2\,+\,5x\,+\,5h\,-\,5\,-\,4x^2\,-\,5x\,+\,5\)
. . . . . . . . . . . . . . . . . . .\(\displaystyle =\:8xh\,+\,4h^2\,+\,5h\) **

\(\displaystyle [3]\;\;\frac{f(x\,+\,h)\,-\,f(x)}{h}\;=\;\frac{8xh\,+\,4h^2\,+\,5h}{h}\;=\;\frac{h(8x\,+\,4h\,+\,5)}{h}\:=\:8x\,+\,4h\,+\,5\)


You see, if you do the three steps separately, it makes more sense ... and it's easier.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** . IMPORTANT!

At this point, <u>two</u> things should have happened:
. . All of \(\displaystyle f(x)\) should have subtracted out.
. . Each remaining term must contain an \(\displaystyle h\).

If this is not true, go back . . . you've made an error!

The most common errors:
. . Squaring incorrectly: \(\displaystyle (x + h)^2\) is NOT \(\displaystyle x^2 + h^2\) !!
. . Subtracting incorrectly: forgetting to "take the minus through".

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I have always taught the Difference Quotient as a three-step recipe.
And I have always pointed out the two Important warning signs.

From the number of related postings, I get the sinking feeling that
. . I'm the only teacher in the Alpha Quadrant that does this.
 
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