Posted by linda on .
State two intervals in which the function y=7sin(1/5x)+2 has an average rate of change that is:
a) zero
b)a negative value
c)a positive value
Please explain how to get the intervals! Thanks a lot in advance!

Math  Trig please help! 
Steve,
a) pick two values of x where y is the same. The average rate of change for that interval will be zero.
For example, if x = pi, then y = 7sin(pi/5)+2
Now sin(4pi/5) = sin(pi/5), so if x = 4pi, then
y = 7sin(4pi/5) + 2
so, on the interval [pi,4pi] the average rate of change is zero.
Taking the easy way out, just tweak the ends of the interval to make the slop of the line negative or positive.
Since sin(pi/5+.01) > sin(pi/5)
the slope on [pi+.05,4pi] will be negative
similarly, on [pi.05,4pi], the slope is positive 
Math  Trig please help! 
kutha kamina,
boi u suck cant even do this question