Question

the function ( f ) is given by ( f(x)=1 + 3cos x ). what is the average rate of change of ( f ) over the interval ( 0,pi )?
a ( -\frac{6}{pi} )
b ( -\frac{2}{pi} )
c ( \frac{2}{pi} )
d 1

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(a = 0\), \(b=\pi\), and \(f(x)=1 + 3\cos x\).

Step2: Calculate \(f(0)\) and \(f(\pi)\)

• For \(x = 0\): \(f(0)=1+3\cos(0)\). Since \(\cos(0)=1\), then \(f(0)=1 + 3\times1=4\).
• For \(x=\pi\): \(f(\pi)=1+3\cos(\pi)\). Since \(\cos(\pi)=- 1\), then \(f(\pi)=1+3\times(-1)=1 - 3=-2\).

Step3: Substitute into the average - rate - of - change formula

\(\frac{f(\pi)-f(0)}{\pi - 0}=\frac{-2 - 4}{\pi}=\frac{-6}{\pi}\)

Answer:

A. \(-\frac{6}{\pi}\)