Mathematics - Application of Derivative Question with Solution | TestHub

$f\left(x\right)=\frac{x^2-3x-6}{x^2+2x+4}$

$f^{\text{'}}\left(x\right)=\frac{\left(x^2+2x+4\right)\left(2x-3\right)-\left(x^2-3x-6\right)\left(2x+2\right)}{{\left(x^2+2x+4\right)}^2}$

$=\frac{5x^2+20x}{{\left(x^2+2x+4\right)}^2}=\frac{5x(x+4)}{{\left(x^2+2x+4\right)}^2}$

Hence, it is increasing in $(-\infty ,-4]\cup [0,\infty )$

Now as $x\to -\infty$ then $f\left(x\right)=1,$also, $x\to \infty$ then $f\left(x\right)=1$

$f\left(-4\right)=\frac{16+12-6}{16-8+4}=\frac{22}{12}=\frac{11}{6}$

$f\left(0\right)=\frac{-3}{2}$

Hence, range is $\left[\frac{-3}{2}, \frac{11}{6}\right]$, and the function is into.