Maxima and Minima — MATHS STD 12 Science — Question

$\text{f}(\text{x})=\frac{\text{x}}{4-\text{x}+\text{x}^{2}}$
$\Rightarrow\text{f}'(\text{x})=\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}$
To find minimum or maximum value $f'(x) = 0$
$\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}=0$
$4 - x^2 = 0$
$\text{x}=\pm2\notin[-1, 1]$
$\text{f}(-1)=\frac{-1}{6}$ and $\text{f}(1)=\frac{1}{4}$
Hence, maximum value is $\frac{1}{4}$.
NOTE: options in the book are not matching with the solution.
Above solution is based on the question given in the book.