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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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    $\begingroup$ Your work for the third problem is correct. $\endgroup$
    – Théophile
    Aug 7, 2012 at 18:43
  • $\begingroup$ @Théophile Great! Thanks for checking it. $\endgroup$ Aug 7, 2012 at 18:44

2 Answers 2

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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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  • $\begingroup$ Does it come out to be $h+6$? $\endgroup$ Aug 7, 2012 at 17:45
  • $\begingroup$ @AustinBroussard No. $\endgroup$
    – M Turgeon
    Aug 7, 2012 at 17:46
  • $\begingroup$ No. I have edited the post to add details. Follow the calculations through, and let me know if you have any other issues :) $\endgroup$
    – Emily
    Aug 7, 2012 at 17:46
  • $\begingroup$ Nevermind. I had a $+$ instead of a $-$. Is it just $h$? $\endgroup$ Aug 7, 2012 at 17:47
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    $\begingroup$ No, it is not just $h$. Read my post more carefully. If $f(x) = 2x+3$, then what happens when you replace $x$ with $x+h$? The answer to that question is the value of $f(x+h)$. $\endgroup$
    – Emily
    Aug 7, 2012 at 17:50
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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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    $\begingroup$ @AustinBroussard Read my edit of my answer for the explanation of how to handle subtraction of fractions with different denominators. It may be helpful to review that material anyhow... in other words, how do you compute $\frac{1}{3}-\frac{1}{7}$... $\endgroup$
    – Emily
    Aug 7, 2012 at 18:05
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    $\begingroup$ For that, wouldn't you multiply $\frac{1}{3}$ by $7$ and $\frac{1}{7}$ by $3$ and then subtract the numerator. $\endgroup$ Aug 7, 2012 at 18:06
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    $\begingroup$ OH. I see it now. $\endgroup$ Aug 7, 2012 at 18:07
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    $\begingroup$ @AustinBroussard No, you would multiply $\frac{1}{3}$ by $\frac{7}{7}$, and so forth, which I think is what you meant. Either way, the same rules hold when subtracting fractions of variables. Try applying those rules and see where it gets you. $\endgroup$
    – Emily
    Aug 7, 2012 at 18:09
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    $\begingroup$ @AustinBroussard You use distributivity; see my edit. $\endgroup$
    – M Turgeon
    Aug 7, 2012 at 18:20

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