Question

The maximum value of f(x) = $\frac{x}{4+x+x^2}$ on [$-$1,1] is

(a) $-\frac{1}{4}$

(b) $-\frac{1}{3}$

(c) $\frac{1}{6}$

(d) $\frac{1}{5}$

Answer 1

(c) $\frac{1}{6}$


$$\begin{gathered} \mathrm{Given}:f\left(x\right)=\frac{x}{4+x+x^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{4+x+x^2-x\left(1+2x\right)}{{\left(4+x+x^2\right)}^2} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{4+x+x^2-x\left(1+2x\right)}{{\left(4+x+x^2\right)}^2}=0 \\ \Rightarrow 4+x+x^2-x\left(1+2x\right)=0 \\ \Rightarrow 4-x^2=0 \\ \Rightarrow x=\pm 2\notin \left[-1,1\right] \\ \mathrm{The}\mathrm{values}\mathrm{of}f\left(x\right)\mathrm{at}\mathrm{extreme}\mathrm{points}\mathrm{are}\mathrm{given}\mathrm{by} \\ f\left(1\right)=\frac{1}{4+1+1^2}=\frac{1}{6} \\ f\left(-1\right)=\frac{-1}{4-1+{\left(-1\right)}^2}=\frac{-1}{4} \\ \mathrm{Thus},\frac{1}{6}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{value}. \end{gathered}$$