# How do you find the average rate of change of f(x) = sec(x) from x=0 to x=pi/4?

Sep 24, 2014

The average rate of change is the slope of the secant line through the points $\left(0 , 1\right)$ and $\left(\frac{\pi}{4} , \sqrt{2}\right)$.

Average rate of change of = $\frac{f \left(b\right) - f \left(a\right)}{b - a}$

$f \left(x\right) = \sec \left(x\right)$

$b = \frac{\pi}{4}$

$a = 0$

$\frac{f \left(b\right) - f \left(a\right)}{b - a} = \frac{f \left(\frac{\pi}{4}\right) - f \left(0\right)}{\frac{\pi}{4} - 0} = \frac{\sec \left(\frac{\pi}{4}\right) - \sec \left(0\right)}{\frac{\pi}{4} - 0}$

Using the knowledge of the Unit Circle we know that $\frac{\pi}{4}$ represents the special triangle $45 , 45 , 90 \to 1 , 1 , \sqrt{2}$

$\sec \left(\frac{\pi}{4}\right) = \frac{h y p}{a \mathrm{dj}} = \frac{\sqrt{2}}{1} = \sqrt{2}$

$\sec \left(0\right) = 1$

$= \frac{\sec \left(\frac{\pi}{4}\right) - \sec \left(0\right)}{\frac{\pi}{4} - 0} = \frac{\sqrt{2} - 1}{\frac{\pi}{4} - 0} = \frac{\sqrt{2} - 1}{\frac{\pi}{4}}$

$= \frac{0.4142}{0.7854} = 0.5274$