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The interval for which the function f(x)=xx2+x+1f(x)=\frac{x}{x^2+x+1} is an increasing function, is:

Solution

✅ Correct Option: 3

f′(x)=(x2+x+1)−x(2x+1)(x2+x+1)2=1−x2(x2+x+1)2f'(x)=\frac{(x^2+x+1)-x(2x+1)}{(x^2+x+1)^2}=\frac{1-x^2}{(x^2+x+1)^2}.

The denominator is always positive, so f′(x)>0f'(x)>0 exactly when 1−x2>01-x^2>0, i.e. on (−1,1)(-1,1).

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