Find the range of the function fx=

1.	Find the range of the function fx=
Answer» $\large\underline{\bold{Given \:Question - }}$ $\sf \: Find \: the \: range \: of \: the \: function \: f(x) = \dfrac{ {x}^{2} + 2x + <klux>1</klux>}{ {x}^{2} - 8x + 12 }$ $\large\underline{\bold{Solution-}}$ Following steps for used finding the range of a function are: Write down y = f(x) and then SOLVE the equation for x, giving something of the FORM x = g(y). Find the domain of g(y), and this will be the range of f(x). Note :- If you can't seem to solve for x, then try graphing the function to find the range. Now, Let's solve the problem now!! STEP :- 1 $\sf \: Let \: f(x) = y \: = \: \dfrac{ {x}^{2} + 2x + 1}{ {x}^{2} - 8x + 12 }$ $\sf \: {yx}^{2} - 8xy + 12y = {x}^{2} + 2x + 1$ $\sf \: {yx}^{2} - 8xy + 12y - {x}^{2} - 2x - 1 = 0$ $\sf \: (y - 1) {x}^{2} - x(8y + 2) + 12y - 1 = 0$ Step :- 2 Discriminant of the above QUADRATIC expression is $\sf \: Discriminant, D = {b}^{2} - 4ac$ $\sf \: D = {(8y + 2)}^{2} - 4(y - 1)(12y - 1)$ $\sf \: = 4 {(4y + 1)}^{2} - 4( {12y}^{2} - y - 12y + 1)$ $\sf \: = 4\bigg( {16y}^{2} + 1 + 8y - 12 {y}^{2} + 13y - 1 \bigg)$ $\sf \: =4( {4y}^{2} + 21y)$ Now, solution of above quadratic equation is given by $\rm :\longmapsto\:\sf \: x \: = \: \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$ $\rm :\longmapsto\:x = \dfrac{8y + 2 \: \pm \: \sqrt{4y(4y + 21)} }{2(y - 1)}$ $\rm :\longmapsto\:x = \dfrac{4y + 1 \: \pm \: \sqrt{y(4y + 21)} }{(y - 1)}$ $\rm :\longmapsto\:y(4y + 21) \geqslant 0$ $\bf\implies \:y \leqslant - \dfrac{21}{4} \: \: \: or \: \: \: y \geqslant 0$

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$\large\underline{\bold{Given \:Question - }}$