# Find the rate of change on an algebraic unknown

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:

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

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?

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

• 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 – Doug Fir Aug 21 '19 at 15:31
• We note that $b-2$ and $2-b$ are negatives and we want to "cancel them out". This fraction does just that. – Ross Millikan Aug 21 '19 at 15:32
• 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 – Doug Fir Aug 21 '19 at 15:39
• 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 – Doug Fir Aug 21 '19 at 15:41

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

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

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

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

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

Is it clear now? Please let me know.

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