- #1

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- Thread starter chukie
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- #1

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- #2

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- #3

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- #4

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ive managed to get the graph

- #5

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i also looked at the mean value theorem. it deals with f'(x)=0 ?

- #6

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Ah, ok. Yes, c=4 since f is a continuous function on [-5,c], then the average value of f from -5 to c = [tex]\frac{1}{c-(-5)}\[ \int_{-5}^{c} f(x)\,dx.\] [/tex]. Since the average value = 0, then it's just the integral from -5 to 4(since it's a graph, yeah the areas cancel out)

No, check out the MVT for integrals also known as the average value of a function.i also looked at the mean value theorem. it deals with f'(x)=0 ?

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- #7

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okay great! thanks so much =)

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