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There is a method in which the given function is like $$\frac{\text{Quadratic}_1}{\text{Quadratic}_2}.$$

The method is to set this equal to $t$ and then cross multiply to get a quadratic equation in $x$ which is variable in quadratic function original one. Then we put discriminant $\geq 0$ to obtain range of $t$.

Doubt

  1. Why we put discriminant $\geq 0$?

  2. Do we have to find domain first and then check for range or directly we can find range of these functions?

Thanks.

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    You need to find the domain of the rational function, because you may need to exclude $x$-values if there is an $x$ that makes your denominator zero. You must make sure your denominator is not zero before you cross-multiply. – 2017-02-05
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    Also, a positive discriminant ensures that you have real-valued answers, so you are asked to find the values of $t$ that ensure that the discriminant is positive. Just treat $t$ as a constant you don't know in this case. – 2017-02-05

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Note that : $ax^2+bx+c=o $ has real root when $\Delta \geq 0$ . Finding domain is important ,because when denominator has root $f(x) \to \infty $ ,But when has no root range of function is in form of $f(x) \in [A,B]$ .look some example $$y=\dfrac{2x^2+2x+4}{x^2+1}\\D_f=(-\infty,+\infty) \to \\ yx^2+y=2x^2+2x+4 \\(y-2)x^2-2x+(y-4)=0 \to \Delta \geq 0\\(-2)^2-4(y-2)(y-4) \geq0 \to (y-2)(y-4) \leq 1 \to\\y \in[1.5,4.41]$$

$$y=\dfrac{x^2+x}{x^2-3}\\ D_f=D_f=(-\infty,+\infty)-\left\{\pm\sqrt3\right\} \\yx^2-3y=x^2+x\\(y-1)x^2-x-3y=0\\\Delta \geq 0 \to (-1)^2-4(y-1)(-3y)\geq 0 \\y \in(-\infty,0.092] \cup [0.9,+\infty)$$ Hope to make asense.