Question

The maximum value of $f\left(x\right)=\frac{x}{1+4x+x^2}$ is ?

• A. $-\frac{1}{4}$
• B. $-\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$\frac{1}{6}$

The given equation is:

$f\left(x\right)=\frac{x}{1+4x+x^2}$

To find the extremum points we differentiate and equate it to zero

$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+4x+x^2)-(4+2x).x}{(1+4x+x^2{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1+4x+x^2-4x-2x^2}{(1+4x+x^2{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=\frac{1-x^2}{(1+4x+x^2{)}^2}$

$f^{\prime }\left(x\right)=0$

$\Rightarrow \frac{1-x^2}{(1+4x+x^2{)}^2}=0$

$\Rightarrow 1-x^2=0$

$\Rightarrow x=\pm 1$

Now to find whether at the critical points we find a maxima or minima we use the second derivative test.

$\Rightarrow f^{\prime \prime }\left(x\right)=\frac{-2x(1+4x+x^2{)}^2-2(1+4x+x^2\left)\right(4+2x\left)\right(1-x^2)}{(1+4x+x^2{)}^4}$

$\Rightarrow f^{\prime \prime }\left(1\right)=\frac{-2x(1+4+1^2{)}^2-2(1+4+1^2\left)\right(4+2\left)\right(1-1^2)}{(1+4+1^2{)}^4}$

$\Rightarrow f^{\prime \prime }\left(1\right)=-\frac{2}{6^2}$

$\Rightarrow f^{\prime \prime }\left(1\right)<0$

$\Rightarrow f^{\prime \prime }(-1)=\frac{-2x(1-4+(-1{)}^2{)}^2-2(1-4+(-1{)}^2\left)\right(4-2\left)\right(1-(-1{)}^2)}{(1-4+(-1{)}^2{)}^4}$

$\Rightarrow f^{\prime \prime }(-1)=\frac{2}{6^2}$

$\Rightarrow f^{\prime \prime }\left(1\right)>0$

Hence we get a maxima at x =1 and a minima at x = -1 according to the second derivative test. Hence the maximum value is f(1) which is:

$\Rightarrow f\left(1\right)=\frac{1}{1+4+1}$

$\Rightarrow f\left(1\right)=\frac{1}{6}$ .....Answer