Question

Find the domain $f\left(x\right)=\frac{x^2+3x+5}{x^2-5x+4}$

Answer 1

$f\left(x\right)=\frac{x^2+3x+5}{x^2-5x+4}$

$=\frac{x^2+3x+5}{x^2-4x-x+4}$

$=\frac{x^2+3x+5}{x(x-4)-1(x-4)}$

$=\frac{x^2+3x+5}{(x-4)(x-1)}$

In real number denominator cannot be zero so $(x-4)(x-1)\ne 0$

So $x\ne 4$ and $x\ne 1$

Hence domain of function will be real no. expect $1$ & $4$

Hence domain $R-\{1,4\}$