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## anonymous one year ago Need help ASAP please! Two students in your class, Tucker and Karly, are disputing a function. Tucker says that for the function, between x = -3 and x = 3, the average rate of change is 0. Karly says that for the function, between x = -3 and x = 3, the graph goes up through a turning point, and then back down. Explain how Tucker and Karly can both be correct, using complete sentences.

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1. anonymous

@hartnn please help me :(

2. anonymous

@Miracrown I'll give medal, please help!

3. anonymous

@nincompoop

4. anonymous

@ganeshie8 please help me!!

5. Rushwr

@abb0t Please help ????

6. Empty

I think a picture could probably help |dw:1439623078585:dw| So this function's average height is about at y=0 and it has a turning point as well between x=-3 and x=3. There are other possible functions as well that could do this too such as a sine function for instance, even though I basically just made kinda like a parabola for the picture. So hopefully you can use this example to come up with your own answers?

7. Rushwr

Thank you @Empty

8. arindameducationusc

nice @Empty

9. Empty

To find the average height of a function over an interval, you sort of do something similar to taking the average of discrete things. Take the sum of the values over the total, but in calculus we turn these sums for the the values into an integral and the total is the interval: $\text{avg} = \frac{\int_a^b f(x) dx}{b-a}$ So here the average height was 0, and I wanted to make a function that looks like what I have here, so I will use $$f(x)=-x^2+a$$ and solve for $$a$$. $0 = \frac{1}{3-(-3)} \int_{-3}^3 -x^2+a dx$ We can simplify this a bit: $0 = \int_0^3 -x^2 +a dx$ $0 = -\frac{3^3}{3} + a3$ $a=3$ So an example of a function that has a turning point on the interval [-3,3] and an average height of zero is $$f(x)=-x^2+3$$. There are really infinitely many other functions that satisfy this situation though, this just happens to be a simple one to construct that meets our requirements here.

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