Question

Find the range of $x^2+\frac{1}{x^2+2}$ by method of substitution?

Answer 1

$x^2+\frac{1}{x^2+2}$

Let $x^2=t$ so,

$y\left(say\right)=t+\frac{1}{t+2}$

$y=\frac{t^2+1+2t}{t+2}$

$\Rightarrow (t+2)y=t^2+1+2t$

$\Rightarrow t^2+1+2t-ty-2y=0$

$\Rightarrow t^2+(2-y)t-2y+1=0$

Above is the Quadratic equation, so for real solution $D\ge 0$

$(2-y{)}^2-4(1\left)\right(-2y+1)\ge 0$

$\Rightarrow 4+y^2-4y+8y-4\ge 0$

$\Rightarrow y^2+4y\ge 0$

$\Rightarrow y(y+4)\ge 0$

Range of the function $y\in (-\infty ,-4]\cup [0,\infty )$