How do I continue from here? ns am unable to find a suitable condition to proceed.

You are watching: X2 3x 1 x2 x 2

Edit: peak of function in the numerator is at $(frac-32, frac-54)$

and that the role in the denominator is at: $(frac-12,frac34)$but the ain"t much helpful, is it?


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$$y=1+frac2xx^2+x+1$$

The duty $y(x)$ is continuous and differentiable in $Bbb R$. Moreover$lim_x opminftyy(x)=1$, therefore $y(x)$ is bounded. This means that the selection of $y$ is an interval $I$.

Then$$y"=frac2x^2+2x+2-4x^2-2x(x^2+x+1)^2=frac2-2x^2(x^2+x+1)^2$$which vanishes at $x=pm1$.

Compute $y(1)$ and $y(-1)$ to find the endpoints the $I$.


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You have shown that $y$ is in the selection of the function$$y = fracx^2 + 3x + 1x + 1$$if $x$ is a real-valued source of the quadratic equation$$x^2(y - 1) + x(y - 3) + (y - 1) = 0$$For $x$ to be a real-valued source of the quadratic equation, that discriminant have to be nonnegative. The discriminant iseginalign*Delta & = b^2 - 4ac\ & = (y - 3)^2 - 4(y - 1)(y - 1)\ & = y^2 - 6y + 9 - 4(y^2 - 2y + 1)\ & = -3y^2 + 2y + 5 endalign*Hence, we require that eginalign*-3y^2 + 2y + 5 & geq 0\3y^2 - 2y - 5 & leq 0\3y^2 + 3y - 5y - 5 & leq 0\3y(y + 1) - 5(y + 1) & leq 0\(3y - 5)(y + 1) & leq 0endalign*which stop if $y in <-1,5/3>$. Therefore, the range of $$y = fracx^2 + 3x + 1x^2 + x + 1$$is $$ extRan = left<-1, frac53 ight>$$


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Points you should examine are whereby $dfracdydx = 0$ and also limits of the range $lim_pminfty dfracdydx$


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$y = 1 +frac2xx^2 + x + 1 = 1 + frac21+x+1/x$, $x e 0.$

A) let $xgt 0:$

1) $x+1/x ge 2$

AM-GM: $x+1/x ge 2sqrtx×1/x$

$y_max = 1 + frac21+2 = 5/3$.

B) permit $x lt 0:$

2) $x+1/x le -2$

$y_min = 1 + frac21- 2 = 1+ (-2) = -1.$

Since

$y = fracx^2 + 3x + 1x^2+x+1$ is constant on $gimpppa.orgbbR$, us find:

Range$ = <-1,5/3>.$


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