What is the maximum and the minimum values of the function,
$f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 2 x + 7}$

4, 2

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2, 1

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1, 0

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None of these

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Solution

The correct option is C 1, 0
Let $y = \frac{x^{2} + 2 x + 1}{x^{2} + 2 x + 7}$
$y = \frac{(x + 1)^{2}}{(x + 1)^{2} + 6}$
$y = \frac{1}{1 + \frac{6}{(x + 1)^{2}}}$
For minimum y, x = -1 then y = 0
For maximum y, x can take an infinitely high value which makes the fraction $\frac{6}{(x + 1)^{2}}$ equal to 0. Hence, maximum value of y is 1.

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What is the maximum and the minimum values of the function, f(x)=x2+2x+1x2+2x+7

The correct option is C 1, 0Let y=x2+2x+1x2+2x+7 y=(x+1)2(x+1)2+6 y=11+6(x+1)2 For minimum y, x = -1 then y = 0 For maximum y, x can take an infinitely high value which makes the fraction 6(x+1)2 equal to 0. Hence, maximum value of y is 1.

What is the maximum and the minimum values of the function,
$f (x) = \frac{x^{2} + 2 x + 1}{x^{2} + 2 x + 7}$