Calculus question need some help
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Calculus question need some help

[From: ] [author: ] [Date: 12-07-16] [Hit: ]
Well, there is only one root in the interval.It is at x=-2.(The other root,So we know that it is always positive but touches the x-axis at x=-2.For f(x) to be decreasing,......
Hi, I have asked this question before but was confused by the explanation. Can someone please break down with steps? The answer was decreasing on inteval - infinity Let f be the function for all real numbers such that f ' (x) = absolute value [x^2 - 4] / [x - 2]
Then f is a decreasing function on the interval ?

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All right, so let's find the roots.

Well, there is only one root in the interval.

It is at x=-2.

(The other root, at x=2 is not included in the interval)


So we know that it is always positive but touches the x-axis at x=-2.

For f(x) to be decreasing, the derivative has to be negative.

But with an absolute value for the derivative, we know it is always positive.



So f(x) is not a decreasing function over those intervals.

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I am not exactly sure how this is decreasing.

If you think about what an absolute value means on a graph, it means that x CAN be negative but y CAN'T be negative.

This meaning if this is the derivative of something and it is f'(x), then f(x) must ALWAYS be increasing since the first derivative is positive.

Therefore, if f'(x) must always be positive, then f(x) must always be increasing.
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