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For fractions A=[tex]\frac{n^{2}+2n+1 }{n^{2}+1 } where n is an integer.Find the value of n so that fraction A has the largest value

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Answer:

To find the value of n that makes the fraction A have the largest value, we can take the derivative of A with respect to n, set it equal to zero, and solve for n.

First, let's rewrite the fraction A as:

A = 1 + \frac{1}{n^{2}+1}

Next, let's take the derivative of A with respect to n:

dA/dn = 0 - 2n/(n^2+1)^2

Setting dA/dn equal to zero, we get:

0 = -2n/(n^2+1)^2

Multiplying both sides by (n^2+1)^2, we get:

0 = -2n

Therefore, n = 0 is a critical point of A. However, n cannot equal 0 because the denominator of A would be zero, which is undefined.

Instead, let's consider the limit of A as n approaches infinity. We have:

lim A = lim (1 + 1/(n^2+1)) = 1

Therefore, as n approaches infinity, A approaches 1.

On the other hand, as n approaches negative infinity, A approaches -1 because the denominator of A becomes negative while the numerator remains positive.

Therefore, the largest value of A occurs at n = infinity, and A approaches 1 as n approaches infinity.

In conclusion, the value of n that makes the fraction A have the largest value is infinity.

User Alexander Jank
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