The maximum value of x2-x-1-x2-x-1 for all real values of x is

The correct option is D 3 f(x) = x^2 x+1x^2+x+1 f'(x) = (x^2+x+1)(2x 1) (x^2 x+1)(2x+1)(x^2+x+1)^2 = 2(x^2 1)(x^2+x+1)^2 For maxima or minima of f ...

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Standard XII

Mathematics

Global Maxima

The maximum v...

Question

The maximum value of $\frac{x^{2} - x + 1}{x^{2} + x + 1}$ for all real values of x is

1/2

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Solution

The correct option is D 3
$f (x) = \frac{x^{2} - x + 1}{x^{2} + x + 1}$

$f^{'} (x) = \frac{(x^{2} + x + 1) (2 x - 1) - (x^{2} - x + 1) (2 x + 1)}{(x^{2} + x + 1)^{2}} = \frac{2 (x^{2} - 1)}{(x^{2} + x + 1)^{2}}$

For maxima or minima of $f$

$f^{'} (x) = 0 \Rightarrow x = \pm 1$

Clearly at $x = - 1$ , $f (x)$ is having maximum value which is $3$ .

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The maximum value of x2−x+1x2+x+1 for all real values of x is

The correct option is D 3f(x)=x2−x+1x2+x+1f′(x)=(x2+x+1)(2x−1)−(x2−x+1)(2x+1)(x2+x+1)2=2(x2−1)(x2+x+1)2For maxima or minima offf′(x)=0⇒x=±1Clearly at x=−1 , f(x) is having maximum value which is 3.

The maximum value of $\frac{x^{2} - x + 1}{x^{2} + x + 1}$ for all real values of x is