Question

Find the range of the function f(x) = x^2+2x+1/x^2-8x+12​

Answer 1

Step-by-step explanation:

Let $\tt\:y=f(x)=\frac{x^2+2x+1}{x^2-8x+12}$

Now,

$\tt\:y( {x}^{2} - 8x + 12)=x^2+2x+1$

$\tt\: \implies \: y {x}^{2} - 8yx + 12y=x^2+2x+1$

$\tt\: \implies \: (y - 1) {x}^{2} - (8y + 2)x + (12y - 1)=0$

Now, its discriminant will be greater than or equal to 0, as x is defined for all real numbers,

$\tt\: \implies \: (8y + 2)^{2} - 4(y - 1)(12y - 1) \geqslant 0$

$\tt\: \implies \: 64y^{2} + 32y + 4 - 4 (12 {y}^{2} - y - 12y + 1) \geqslant 0$

$\tt\: \implies \: 64y^{2} + 32y + 4 - 48 {y}^{2} + 52y - 4\geqslant 0$

$\tt\: \implies \: 16y^{2} + 84y \geqslant 0$

$\tt\: \implies \: 4y(4y + 21) \geqslant 0$

$\tt\: \implies \: 16y \bigg(y + \frac{21}{4} \bigg) \geqslant 0$

$\tt\: \implies \: y \bigg(y + \frac{21}{4} \bigg) \geqslant 0$

$\tt \pink{\implies \: y \in \bigg( - \infty , - \frac{21}{4} \bigg] \: \: \cup \: \bigg[ 0 , \infty \bigg) }$