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I am trying to study for a test I have tomorrow but I can't even interpret what is happening in my book.

Find an equation of the tangent line at the hyperbola $y=\dfrac3{x}$ at the point $(3,1)$

Let $f(x) = \dfrac3{x}$ then the slope of the tangent at $(3,1)$ is:

$$m = \lim _{h \to 0} \frac{f(3+h)-f(3)}{h} = \lim _{h \to 0} \frac{\frac{3}{3+h} - 1}{h} = \lim _{h \to 0} \frac{\frac{3-(3+h)}{3+h}}{h} = \lim _{h \to 0} -\frac{h}{h(3+h)} = \lim _{h \to 0} \frac{-1}{3+h} = -\frac{1}{3}$$

I do not follow what happened from the first to second step, they somehow manipulate the problem to make the number into the denominator over 3. Not sure how to that but it does need seem like good algebra especially considering h was not in any way manipulated which should change the problem.

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    $\dfrac{\frac3{3+h}-1}{h}=\dfrac{\frac{3-3-h}{3+h}}{h}=\frac{-h}{h(3+h)}$... can you see what cancels?2011-09-11
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    I don't even see what happened between the two problems. It doesn't seem possible with proper algebra.2011-09-11
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    Which two problems?2011-09-11
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    Never mind I got it.2011-09-11
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    Your parentheses need work. Just after $0$ they are mismatched. After the second $=$ they are not correct, which may be your problem. The second $3+h$ is the denominator for all that precedes it.2011-09-11
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    The $-1$ became $-\frac{3+h}{3+h}$, then the preceding $3$ was put over the same denominator.2011-09-11
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    Well it isn't my problem. I am reading out of the book. I may have typed it out wrong though. Looking at it again I see now problems with it. At least nothing that should cause confusion. Still not sure what happens to make it -h...2011-09-11
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    What did you expect $3-3-h$ to end up being?2011-09-11
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    Ohh I think I just got it F(3) is -1, there was no manipulation and f(3+h) is some crazy math that results in 3/(3+h) somehow 3-3-h to me looks like it is -h.2011-09-11
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    No, $f(3)=3/3=1$... and nothing crazy about $f(3+h)$; all that was done is to *replace* the $x$ in $3/x$ with $3+h$, nothing more.2011-09-11
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    What happens with f(3+h) I am not sure how to manipulate these problems. I know it is the function of 3 and the function oh h added togther. How do I get that x = 3/x?2011-09-11
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    Your $f(x)$ was *defined* to be $\frac3{x}$. You know, the line that goes "Let..." Then to do $f(3+h)$, the $x$ in $\frac3{x}$ is replaced with $3+h$. That's why you get $\frac3{3+h}$.2011-09-11
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    I don't get why it is replaced with 3+h.2011-09-11

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From the first to second step we used the definition of $f$ to evaluate $f(3+h)$ and $f(3)$. From the second to the third we used $-1=-\frac{3+h}{3+h}$ then combined the two fractions in the numerator over the common denominator. From the third to the fourth the $3$ and $-3$ are cancelled and numerator and denominator are multiplied by $\frac{1}{h}$. Then the $h$'s are cancelled (as $h \ne 0$) and the limit is taken.