What are the absolute extrema of #f(x)=(x3)/(x^2+x7) in(0,5)#?
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To find the absolute extrema of the function (f(x) = \frac{x3}{x^2+x7}) on the interval ((0, 5)), we follow these steps:
 Find critical points by setting the derivative of (f(x)) equal to zero and solving for (x).
 Evaluate (f(x)) at these critical points as well as at the endpoints of the interval.
 Determine which of these values is the maximum and which is the minimum.
Let's begin by finding the critical points:

Find the derivative of (f(x)) using the quotient rule: [f'(x) = \frac{(x^2+x7)(1)  (x3)(2x+1)}{(x^2+x7)^2}]

Set (f'(x)) equal to zero and solve for (x): [\frac{(x^2+x7)(1)  (x3)(2x+1)}{(x^2+x7)^2} = 0] [x^3  3x^2  6x + 22 = 0]
This cubic equation may not have simple solutions, so we'll use numerical methods or calculus software to approximate the critical points.

Evaluate (f(x)) at these critical points as well as at the endpoints (x = 0) and (x = 5). [f(0), f(\text{critical points}), f(5)]

Identify the maximum and minimum values among these.
Once the critical points and the function values at the endpoints are determined, we can compare them to find the absolute extrema within the given interval ((0, 5)).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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