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I know that the average rate of change formula is: $$\frac{f(b) - f(a)}{b-a},$$ but I'm not sure if my arithmetic is current for this problem: Find the average rate of change on the interval $(1/100,2/100)$. Let $f(x) =1/x$.

Here is how I did it so far: $$\frac{\frac{1}{\frac{2}{100}} - \frac{1}{\frac{1}{100}}}{\frac{2}{100}- \frac{1}{100}}$$

I'm not sure if I'm applying the numbers currently. What should I get as my final result?

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  • $\begingroup$ You are missing brackets in your numerator. This tutorial explains how to typeset mathematics on this site. If you write $\frac{f(b) - f(a)}{b - a}$, you will obtain $\frac{f(b) - f(a)}{b - a}$. If you write $$\frac{f(b) - f(a)}{b - a}$$, you will obtain $$\frac{f(b) - f(a)}{b - a}$$ where the fraction is displayed on its own line. $\endgroup$ – N. F. Taussig Oct 27 '18 at 16:45
  • $\begingroup$ Thanks! I didn't know about that! I'm just kind of struggling with this math problem 😅 $\endgroup$ – User231 Oct 27 '18 at 16:51
  • $\begingroup$ In the numerator, remember that when you divide by a fraction, you multiply by its reciprocal, so $\frac{1}{\frac{1}{100}} =1 \cdot \frac{100}{1} = 100$. $\endgroup$ – N. F. Taussig Oct 27 '18 at 17:05
  • $\begingroup$ do you mind showing the whole problem? I need to make sure that my solutions for the denominator are correct. $\endgroup$ – User231 Oct 30 '18 at 1:48
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We are given the function $f(x) = \frac{1}{x}$ and asked to calculate its average rate of change over the interval $(\frac{1}{100}, \frac{2}{100})$. \begin{align*} \frac{f\left(\frac{2}{100}\right) - f\left(\frac{1}{100}\right)}{\frac{2}{100} - \frac{1}{100}} & = \frac{\frac{1}{\frac{2}{100}} - \frac{1}{\frac{1}{100}}}{\frac{2}{100} - \frac{1}{100}} \tag{1}\\ & = \frac{1 \cdot \frac{100}{2} - 1 \cdot \frac{100}{1}}{\frac{1}{100}} \tag{2}\\ & = \frac{50 - 100}{\frac{1}{100}} \tag{3}\\ & = \frac{-50}{\frac{1}{100}} \tag{4}\\ & = -50 \cdot \frac{100}{1} \tag{5}\\ & = -5000 \tag{6} \end{align*}

(1) Substitute into the formula $$\frac{f(b) - f(a)}{b - a}$$ when $a = \frac{1}{100}$, $b = \frac{2}{100}$ and $f(x) = \dfrac{1}{x}$.

(2) In the numerator, dividing by a fraction is equivalent to multiplying by its reciprocal. To see this, observe that if $a, b, c, d$ are integers with $b, c, d \neq 0$, then $$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{\frac{a}{b}}{\frac{c}{d}} \cdot \frac{\frac{d}{c}}{\frac{d}{c}} = \frac{\frac{ad}{bc}}{1} = \frac{ad}{bc} = \frac{a}{b} \cdot \frac{d}{c}$$ In the denominator, the two fractions have a common denominator, so we may subtract them. If $c \neq 0$, then
$$\frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}$$

(3) Since $1$ is the multiplicative identity, $$1 \cdot \frac{100}{2} - 1 \cdot \frac{100}{1} = \frac{100}{2} - \frac{100}{1} = 50 - 100$$

(4) Subtract the integers in the numerator.

(5) To divide by a fraction, multiply by its reciprocal.

(6) Since $1$ is the multiplicative identity, $$-50 \cdot \frac{100}{1} = -50 \cdot 100 = -5000$$

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