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Let $f(x) = \frac{1}{x}$. Find the number b such that the average rate of change of $f$ on the interval $(2, b)$ is $-\frac{1}{10}$

Yesterday I attempted this question by calculating $f(x_1)$ and $f(x_2)$:

$f(x_1)$ = $\frac{2}{4}$ = $\frac{1}{2}$

$f(x_2)$ = $\frac{1}{b}$

Change is $\frac{f(x_2)-f(x1)}{b - 2}$

So:

$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$

Yesterday I asked on here how to correctly subtract the fraction in the numerator.

I was able to get as far as rewriting the fraction in the numerator on my own as:

$\frac{\frac{2-b}{2b}}{b-2}$

The solution in the question that I posted went further and managed to simplify this to $-\frac{1}{2b}$

I did not really understand how this was arrived at and I was hoping that someone could 'hold my hand' to understand how this was arrived that? That is my first question.

For the second part I was able to complete myself: Calculate $b$ such that the rate of change on the interval $(2,b)$ is $-\frac{1}{10}$:

$-\frac{1}{2b} = -\frac{1}{10}$

$\frac{1}{2b} = \frac{1}{10}$ # multiply both sides by -1

$1 = \frac{2b}{10}$ # multiply out denominator on left side so multiply both sides by 2b

$10 = 2b$ # multiply out denominator on right side, multiply both sides by 10

$b = 5$ #tada

It is the in between step that I am confused about. I do not really follow how to go from this:

$\frac{\frac{2-b}{2b}}{b-2}$

To this:

$-\frac{1}{2b}$

How exactly was that done? In baby steps if possible?

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When you have a fraction, you often simplify it by multiplying it by $1$ in the form of $\frac aa$ for some convenient $a$. Here you have a fraction with $ \frac {2-b}{2b}$ in the numerator and $b-2$ in the denominator. A convenient $a$ is $\frac 1{b-2}$. Then we have $$\frac {\frac {2-b}{2b}}{b-2}=\frac {\frac {2-b}{2b}}{b-2}\cdot \dfrac{\left(\frac 1{b-2}\right)}{\left(\frac 1{b-2}\right)}\\ =\frac {\frac {2-b}{2b}\cdot {\frac 1{b-2}}}{(b-2)\frac 1{b-2}}\\=\frac{\left(\frac{-1}{2b}\right)}1\\=\frac{-1}{2b}$$

Another common approach is $$\frac {\left({\frac ab}\right)}{\left({\frac cd}\right)}=\frac {\left({\frac ab}\right)}{\left({\frac cd}\right)}\cdot \frac {\left({\frac dc}\right)}{\left({\frac dc}\right)}=\frac {ad}{bc}$$

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  • $\begingroup$ Just trying to wrap my head around this. Why is $\frac{1}{b-2}$ a convenient a? What are we trying to do here by multiplying by a/a? I cannot see the bigger picture $\endgroup$ – Doug Fir Aug 21 at 15:31
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    $\begingroup$ We note that $b-2$ and $2-b$ are negatives and we want to "cancel them out". This fraction does just that. $\endgroup$ – Ross Millikan Aug 21 at 15:32
  • $\begingroup$ So we want to cancel out the numerator in the numerator 2-b? Would it have also worked to start by trying to cancel out the denominator b-2 first? Perhaps by multiplying the whole thing by b-2? Somehow that just 'feels' easier for me to follow but I cannot explain why or if it's a dead end? But my impulse is to multiply out the denominator $\endgroup$ – Doug Fir Aug 21 at 15:39
  • $\begingroup$ Aha! I see your edit. Yes, for some reason I can follow this easier, the second approach of multiplying out the denominator with it's reciprocal $\endgroup$ – Doug Fir Aug 21 at 15:41
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$\displaystyle \frac{\frac{2-b}{2b}}{b-2}$

$\displaystyle = \left(\frac{2-b}{2b} \right)\left(\frac{1}{b-2}\right)$

$\displaystyle = \left(\frac{(-1)(b-2)}{2b} \right)\frac{1}{(b-2)}$

Now cancel $(b-2)$ from the numerator and the denominator to get

$\displaystyle -\frac{1}{2b} $

Is it clear now? Please let me know.

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From the denominator of the fraction $\frac{\frac{2-b}{2b}}{b-2}$ pick up $-1$ and obtain: $ \frac{\frac{2-b}{2b}}{-(2-b)}$. Now rewrite the fraction as: $\frac{2-b}{2b}\cdot \frac{-1}{2-b}$. $2-b$ simplifies and you have: $-\frac{1}{2b}$.

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