Question

Mark the correct alternate of the following
Find the maximum value of $f(x) = \dfrac{x}{{4 + x + {x^2}}}$ on $[ - 1,1]$ is
$\
  A)\dfrac{{ - 1}}{4} \\
  B)\dfrac{{ - 1}}{3} \\
  C)\dfrac{1}{6} \\
  D)\dfrac{1}{5} \\
\ $

Answer 1

Hint: Here to find the maximum value of the function we will differentiate the given function. Apply $\dfrac{{du}}{{dv}}$ rule for differentiation.

Complete Step-by-Step solution:
Given function is $f(x) = \dfrac{x}{{4 + x + {x^2}}}$
Now on differentiating the given function, we get
 So, for differentiation apply $\dfrac{{du}}{{dv}}$ rule for the given function.
 We know that $\dfrac{{du}}{{dv}}$=$\dfrac{{v\dfrac{d}{{dx}}(u) - u\dfrac{d}{{dx}}(v)}}{{{v^2}}}$
$ \Rightarrow f'(x) = \dfrac{{\dfrac{d}{{dx}}(x)(4 + x + {x^2}) - \dfrac{d}{{dx}}(4 + x + {x^2})x}}{{{{(4 + x + {x^2})}^2}}}$
$ \Rightarrow f'(x) = \dfrac{{ - {x^2} + 4}}{{{{(4 + x + {x^2})}^2}}}$
To get the maximum value of the function we know that$f'(x) = 0$
$\
   \Rightarrow \dfrac{{ - {x^2} + 4}}{{{{(4 + x + {x^2})}^2}}} = 0 \\
   \Rightarrow - {x^2} + 4 = 0 \\
   \Rightarrow {x^2} = 4 \\
  \therefore x = \pm 2 \\
\ $
From this we can say that the maximum and minimum value exists $ \pm 2 $
But the given limit is $[ - 1,1]$ where the value $ \pm 2$ does not lie in the limit, so let us make use of given limits.
 Let us substitute the limit values in the given function, then
$f( - 1) = \dfrac{{ - 1}}{4}$
$f(1) = \dfrac{1}{6}$
Hence the maximum value of the function is $f(1) = \dfrac{1}{6}$
Option C is the correct one.

Note: In this problem limits will be neglected which affects the answer. So after finding the maximum (x) value check whether the maximum(x) value lies in the limit or not.