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

Question

Paisley Johnson · Accepted Answer

The average rate of change is the slope of the secant line through the points #(0,1)# and #(pi/4,sqrt(2))#. Average rate of change of = #(f(b)-f(a))/(b-a)# #f(x)=sec(x)# #b=pi/4# #a=0# #(f(b)-f(a))/(b-a)=(f(pi/4)-f(0))/(pi/4-0)=(sec(pi/4)-sec(0))/(pi/4-0)# Using the knowledge of the Unit Circle we know that #pi/4# represents the special triangle #45, 45, 90 -> 1, 1, sqrt(2)# #sec(pi/4)=(hyp)/(adj)=sqrt(2)/1=sqrt(2)# #sec(0)=1# #=(sec(pi/4)-sec(0))/(pi/4-0)=(sqrt(2)-1)/(pi/4-0)=(sqrt(2)-1)/(pi/4)# #=0.4142/0.7854=0.5274#

Paisley Johnson · Answer

undefined To find the average rate of change of $ f(x) = \sec(x) $ from $ x = 0 $ to $ x = \frac{\pi}{4} $, you first need to calculate the values of the function at both endpoints and then use the formula for average rate of change:

$$ \text{Average Rate of Change} = \frac{f(\text{end}) - f(\text{start})}{\text{end} - \text{start}} $$

1. Calculate $ f(0) $ and $ f\left(\frac{\pi}{4}\right) $:
   - $ f(0) = \sec(0) = \frac{1}{\cos(0)} = 1 $
   - $ f\left(\frac{\pi}{4}\right) = \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \sqrt{2} $

2. Plug the values into the formula:
   $$ \text{Average Rate of Change} = \frac{\sqrt{2} - 1}{\frac{\pi}{4} - 0} $$

3. Simplify:
   $$ \text{Average Rate of Change} = \frac{\sqrt{2} - 1}{\frac{\pi}{4}} \times \frac{4}{4} = \frac{4(\sqrt{2} - 1)}{\pi} $$