Range of the function f(x) = frac{x^2+x+2}{x^ | vedclass

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Question

(C) Let $y = \frac{x^2+x+2}{x^2+x+1}$.$y(x^2+x+1) = x^2+x+2$$(y-1)x^2 + (y-1)x + (y-2) = 0$.For $x \in R$,the discriminant $D \geq 0$.$D = (y-1)^2 - 4

Vedclass · Accepted Answer

$\left(1, \frac{7}{3}\right]$

Vedclass · Answer

(C) Let $y = \frac{x^2+x+2}{x^2+x+1}$.$y(x^2+x+1) = x^2+x+2$$(y-1)x^2 + (y-1)x + (y-2) = 0$.For $x \in R$,the discriminant $D \geq 0$.$D = (y-1)^2 - 4(y-1)(y-2) \geq 0$.$(y-1)[(y-1) - 4(y-2)] \geq 0$.$(y-1)(-3y+7) \geq 0$.$(y-1)(3y-7) \leq 0$.Thus,$1 Since $x^2+x+1 = (x+1/2)^2 + 3/4 > 0$,$y$ can never be $1$ because the numerator $x^2+x+2$ is always greater than the denominator $x^2+x+1$.Therefore,the range is $\left(1, \frac{7}{3}\right]$.

Range of the function $f(x) = \frac{x^2+x+2}{x^2+x+1}, x \in R$ is

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