Posted by **linda** on Monday, December 5, 2011 at 11:33pm.

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**, Tuesday, December 6, 2011 at 10:41am
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

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