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Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$ where
a) $f(x)=2x+3,$
b) $f(x)=\frac{1}{x+1},$
c) $f(x)=x^2.$

I believe that if anyone can help me out with the first one, the other two might come clearer to me. But I started out this problem by plugging in $f(x)$ and got: $\dfrac{(2x+3)+f(h)-(2x+3)}{h}$ I have no idea what to do after this. Because the $2x+3$'s can cancel out and leave me with just $\frac{f(h)}{h}$ but that doesn't make sense to me. Please help.

EDIT: The first one is solved and now for the second one, this is what I got: $\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}$ Now, to subtract fractions the denominator has to be the same and they are the exact same except for the first fraction has an $h$ while the other one does not. How would I go about this?

For the third problem, this is all of my work: $\begin{align*}\dfrac{(x+h)^2-x^2}{h}&= \dfrac{x^2+2hx+h^2-x^2}{h}\\ &= \dfrac{2hx+h^2}{h}\\ &= \dfrac{h(2x+h)}{h}\\ &= 2x+h\end{align*}$ Is this all correct?

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    @Théophile Great! Thanks for checking it.2012-08-07

3 Answers 3

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for the first one 2x +3 2 ( X + H ) - ( 2X +3 ) ---------------------------- H 2X + 2 H - 2X + 3 _____________ H
2H =
__ = 2 H

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$f(x+h)$ means replace $x$ with $x+h$ in your function definition. If $f(x) = 2x+3$, then $f(x+h) = 2(x+h)+3$, not $2x+3+f(h)$.

Therefore, $\frac{f(x+h)-f(x)}{h} = \frac{2(x+h)+3-\left(2x+3\right)}{h} = \frac{2x+2h+3-2x-3}{h} = \cdots$

Edit: To answer your second question, how do you handle just $\frac{1}{x+h+1}-\frac{1}{x+1}$? One simple way: multiply the first term above and below by the second term's denominator, and vice-versa: $\frac{1}{x+h+1}\frac{x+1}{x+1}-\frac{1}{x+1}\frac{x+h+1}{x+h+1} = \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}.$

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    I got the first one, I now need help with the second one.2012-08-07
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For the second one, you do the same thing: $\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}=\dfrac{1}{h}\left(\frac{1}{(x+h)+1}-\frac{1}{x+1}\right)=\dfrac{1}{h}\dfrac{(x+1)-(x+h+1)}{(x+1)(x+h+1)}=\dfrac{1}{h}\dfrac{-h}{(x+1)(x+h+1)}=\ldots$

Added: Now, you use the distributivity: $(x+1)(x+h+1)=(x+1)x+(x+1)h+(x+1)1=(x^2+x)+(xh+h)+(x+1)=\ldots$

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    I will re-edit once again for the third problem after I work it out.2012-08-07