Question

The range of \(f(x)=\dfrac1{-x^2+4x+5}\),\(x\in \mathbb R-\{-1,5\}\) is

Answer 1

Let \(y=f(x)\)
\(\therefore y=\dfrac1{-x^2+4x+5}\\
\Rightarrow yx^2-(4y)x-5y+1=0\)
Here, \(y\neq0\) because in quadratic equation coefficient of \(x^2\) can not be \(0.\)

For \(x\in \mathbb R-\{-1,5\}\)
\(D\ge0\\
\Rightarrow 16y^2-4y(1-5y)\ge0\\
\Rightarrow 16y^2-4y+20y^2\ge0\\
\Rightarrow 9y^2-y\ge0\\
\Rightarrow y(9y-1)\ge0\\
\Rightarrow y\in (-\infty,0]\cup\left[\dfrac19,\infty\right)\)
But $y\ne0$
\(\Rightarrow y\in(-\infty,0)\cup\left[\dfrac19,\infty\right)\)