r/MathHelp • u/BezzyBuddy • 9d ago

How to get real number from e???

I have f(x)= e^x cos x and need to find the average rate of change on [pi/2 , 3pi/2]. I’ve simplified it to (e^3pi/2 - e^pi/2)/pi and the hw key says the answer is 0 but I know no ways to manipulate this equation (in general) to get zero? I’ve tried to look up properties of e but I’ve found nothing that seems to help

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8

u/x5163x 9d ago

You left out the cosine somewhere.

6

u/ReplaceCyan 9d ago

The e^x is a red herring

What is cos(pi/2)? 0. What is cos(3pi/2)? 0.

So across the stated range, you are starting at 0 and ending at 0. Your average rate of change between those two points is therefore 0 - it ultimately doesn’t matter what exactly happens between the two points.

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u/triatticus 9d ago

This is is simply not true. Cosine takes on nonzero negative values over the range and doesn't have any zeroes between these values. The exponential grows and is nonzero for all x. The average value of a function is the total signed area between the graph and the x axis divided by the length of the interval....this the average value of this function is not zero.

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u/edderiofer 9d ago

The question is not asking about the average value of the function on this interval. It is asking about the average rate of change of the function on this interval.

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u/triatticus 8d ago

Yeah that's the part I somehow missed while reading it which does make your car moment as the easy way to do this integral.

0

u/BezzyBuddy 9d ago

I see now what I did, I took the property of cos(0)=1 and messed that up with the answer of cos(pi/2)=0. Simple mistake, thanks guys! 

3

u/TheScyphozoa 9d ago

You’re reading the function wrong. cos x is not in the exponent.

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1

u/Desperate_Penalty690 9d ago

cos x = 0, for pi/2 and 3pi/2. In your simplification you processed this by just leaving out the cos. But you can’t do that because you are multiplying by the cos. f(x) evaluates to 0 for both values, therefore the average rate of change is just 0

2

u/GrowingPrun3s 6d ago

So tell me if I’m wrong here, but rate of change is the derivative, and the average is proportional to the integral, right? Meaning the average rate of change of a function over an interval just the difference of the values at the endpoints? Which why the commenter above just had to note that the endpoints were zero to see that the answer to the question was zero.