How do I find the average rate of change of $f (x) = sec x$ $f(x) = sec x$ from 0 to $\frac{π}{4}$ $pi/4$ ?

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How do I find the average rate of change of $f (x) = sec x$ $f(x) = sec x$ from 0 to $\frac{π}{4}$ $pi/4$ ?

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Answer 1 · 2014-09-19T01:12:57

Average rate of change = $\frac{f (b) - f (a)}{b - a}$ $(f(b)-f(a))/(b-a)$ , where $b$ $b$ is the upper bound and $a$ $a$ is the lower bound.

$f (x) = sec x$ $f(x) = sec x$

$\frac{f (\frac{π}{4}) - f (0)}{\frac{π}{4} - 0} = \frac{sec (\frac{π}{4}) - sec (0)}{\frac{π}{4}} = \frac{\sqrt{2} - 1}{\frac{π}{4}} = \frac{0.4142}{0.7854} = 0.5274$ $(f(pi/4)-f(0))/(pi/4-0)=(sec(pi/4)-sec(0))/(pi/4)=(sqrt(2)-1)/(pi/4)=0.4142/0.7854=0.5274$

Things to remember:

Review the unit circle

$\frac{π}{4} = 45$ $pi/4= 45$ , degrees and is a special triangle, $45, 45, 90 : 1, 1, \sqrt{2}$ $45,45,90 : 1,1,sqrt(2)$

$cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $cos (pi/4)=1/sqrt(2)$

$sec (\frac{π}{4}) = \frac{1}{cos (\frac{π}{4})} = \frac{1}{\frac{1}{\sqrt{2}}} = \frac{1}{1} \cdot \frac{\sqrt{2}}{1} = \frac{\sqrt{2}}{1} = \sqrt{2}$ $sec (pi/4)=1/(cos (pi/4))=1/(1/sqrt(2))=1/1*sqrt(2)/1=sqrt(2)/1=sqrt(2)$

$cos (0) = 1$ $cos(0)=1$

$sec (0) = \frac{1}{cos (0)} = \frac{1}{1} = 1$ $sec (0)=1/(cos (0))=1/(1)=1$

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How do I find the average rate of change of $f (x) = sec x$ $f(x) = sec x$ from 0 to $\frac{π}{4}$ $pi/4$ ?

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