Question

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

Answer 1

Hint:A derivative basically finds the slope of a function. First, we need to differentiate the given function. Let \[{f^1}(x) = 0\] and we will get critical numbers. And, then we need to find the second derivative and apply these critical numbers to it. To find the maximum and minimum value we need to apply those x values in the original function.

Complete step by step answer:
Given the function as below,
\[f(x) = \dfrac{x}{{4 + x + {x^2}}}\]and the range is given as\[[ - 1,1]\].
Differentiating the given function w.r.t x, we get,
\[{f^1}(x) = \dfrac{{(4 + x + {x^2})\dfrac{{dx}}{{dx}} - x\dfrac{d}{{dx}}(4 + x + {x^2})}}{{{{(4 + x + {x^2})}^2}}}\]
Since, \[{f^1}(x) = 0\]. We will use this and we will get,
\[ \Rightarrow \dfrac{{(4 + x + {x^2})\dfrac{{dx}}{{dx}} - x\dfrac{d}{{dx}}(4 + x + {x^2})}}{{{{(4 + x + {x^2})}^2}}} = 0\]
\[ \Rightarrow 4 + x + {x^2} - x(1 + 2x) = 0\]
\[ \Rightarrow {x^2} + x + 4 - 2{x^2} - x = 0\]
\[ \Rightarrow {x^2} - 2{x^2} + 4 = 0\]
\[ \Rightarrow - {x^2} + 4 = 0\]
\[ \Rightarrow 4 = {x^2}\]
\[ \Rightarrow {x^2} = 4\]
\[ \Rightarrow x = \pm 2\]

So, we have two values of x.
-2x.
i.e. -2<-1 and 2>1
To distinguish the values of x as the point of maximum or minimum, we need a second derivative of the function.
\[{f^1}(x) = \dfrac{{ - {x^2} + 4}}{{{{({x^2} + x + 4)}^2}}}\]
\[\Rightarrow {f^{11}}(x) = \dfrac{{{{({x^2} + x + 4)}^2}\dfrac{d}{{dx}}( - {x^2} + 4) - ( - {x^2} + 4)\dfrac{d}{{dx}}{{({x^2} + x + 4)}^2}}}{{{{({x^2} + x + 4)}^3}}}\]
\[\Rightarrow {f^{11}}(x) = \dfrac{{{{({x^2} + x + 4)}^2}( - 2x) + ({x^2} + 4)(2)({x^2} + x + 4)(2x + 1)}}{{{{({x^2} + x + 4)}^3}}}\]
Divide numerator and denominator by\[({x^2} + x + 4)\], we will get,
\[{f^{11}}(x) = \dfrac{{({x^2} + x + 4)( - 2x) + ({x^2} + 4)(4x + 2)}}{{{{({x^2} + x + 4)}^2}}}\]
\[\Rightarrow {f^{11}}(x) = \dfrac{{({x^2} + x + 4)( - 2x) + ({x^2} + 4)(4x + 2)}}{{{{({x^2} + x + 4)}^2}}}\]

Substituting the value of\[x = - 1\], we will get,
\[{f^{11}}( - 1) = ({( - 1)^2} + ( - 1) + 4)( - 2( - 1)) + ({( - 1)^2} + 4)(4( - 1) + 2)\]
\[\Rightarrow {f^{11}}( - 1) = (1 - 1 + 4)(2) + (1 - 4)( - 4 + 2) \\
\Rightarrow {f^{11}}( - 1) = 8 + ( - 3)( - 2) \\ \]
Removing the brackets we will get,
\[{f^{11}}( - 1) = 8 + 6 \\
\Rightarrow {f^{11}}( - 1) = 14 \\ \]
Thus, \[{f^{11}}( - 1) = 14\]>\[0\].
This means that \[x = - 1\]is a point of minimum.

Next, substituting the value of\[x = 1\], we will get,
\[{f^{11}}(1) = ({(1)^2} + (1) + 4)( - 2(1)) + ({(1)^2} + 4)(4(1) + 2)\]
\[\Rightarrow {f^{11}}(1) = (1 + 1 + 4)( - 2) + (1 - 4)(4 + 2) \\
\Rightarrow {f^{11}}(1) = 6( - 2) + ( - 3)6 \\ \]
Removing the brackets we will get,
\[{f^{11}}(1) = - 12 - 18 = - 30 \]
Thus, \[{f^{11}}(1) = - 30\]< \[0\]. This means that \[x = 1\] is a point of maximum.
Now, we will find the maximum value as below,
\[f(x) = \dfrac{x}{{4 + x + {x^2}}}\]
Substituting the value of\[x = 1\], we will get,
\[f(x) = \dfrac{1}{{4 + 1 + {1^2}}}\]
\[\Rightarrow f(x) = \dfrac{1}{{4 + 1 + 1}} \\
\therefore f(x) = \dfrac{1}{6} \\ \]
Hence, the maximum value of \[f(x) = \dfrac{x}{{4 + x + {x^2}}}\] for a given range \[[ - 1,1]\] is \[\dfrac{1}{6}\].

Note:The value of the function at a maximum point is called the maximum value of the function and the value of the function at a minimum point is called the minimum value of the function. Then find the second derivative. The function f (x) is maximum when f''(x) < 0 and is minimum when f''(x) > 0. Second Derivative (f''(x)): less than 0 is a maximum, greater than 0 is a minimum and also equal to 0, then the test fails (there may be other ways of finding out though).