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At what rate is this function increasing? |
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| Apr13-10, 10:34 PM | #1 |
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At what rate is this function increasing?
When x = 16,the rate at which [tex]\sqrt x[/tex] is increasing is [tex]\frac {1}{k}[/tex] times the rate at which x is increasing. What is the value of k?
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| Apr13-10, 10:36 PM | #2 |
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I thought it would be 4 but the answer is 8.
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| Apr13-10, 10:42 PM | #3 |
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Its the ratio of the derivatives evaluated at x=16,
(2 sqrt(x))^-1 |
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