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

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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?

\$\$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\$.

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 iseginalign*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>\$\$

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

\$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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