Question

If \[f(x) = \dfrac{1}{{\left[ {4{x^2} + 2x + 1} \right]}}\], then its maximum value is
A. \[\dfrac{4}{3}\]
B. \[\dfrac{2}{3}\]
C. \[1\]
D. \[\dfrac{3}{4}\]

Answer 1

Hint: The standard equation and the coefficient of \[x\] square can be used to determine whether the aforementioned equation has a minimum value or not. If the coefficient of \[{x^2}\] is positive, the equation has a positive value greater than zero, and vice versa. We will resolve the presented issue based on the aforementioned standards.

Complete step by step Solution:
Let's go over the prerequisite for determining the minimal value in greater depth.
The conventional quadratic equation is stated as, and we will compare it to the one given
\[a{x^2} + bx + c = 0\]
We have been given the equation that,
\[\dfrac{1}{{4{x^2}+2x + 1 = 0}}\]
\[f(x) = \dfrac{1}{{4{x^2} + 2x + 1}}\]--- (1)
Now, we have to differentiate the above equation:
Apply exponent rule \[\dfrac{1}{a} = {a^{ - 1}}\]:
\[ = \dfrac{d}{{dx}}\left( {{{\left( {4{x^2} + 2x + 1} \right)}^{ - 1}}} \right)\]
Now, let’s apply chain rule for the above equation, we get
\[ = - \dfrac{1}{{{{\left( {4{x^2} + 2x + 1} \right)}^2}}}\dfrac{d}{{dx}}\left( {4{x^2} + 2x + 1} \right)\]
\[ = - \dfrac{1}{{{{\left( {4{x^2} + 2x + 1} \right)}^2}}}\left( {8x + 2} \right)\]
Now, we have to simplify the above equation, we obtain
\[ \Rightarrow {f^\prime }(x) = \dfrac{{ - (8x + 2)}}{{{{\left( {4{x^2} + 2x + 1} \right)}^2}}}\]
For maxima or minima, put \[{f^\prime }({\rm{x}}) = 0\]
\[ \Rightarrow 8{\rm{x}} + 2{\rm{x}} = - \dfrac{1}{4}\]
Now, we have to again differentiate the above equation that is to find second order derivative, we get
\[{f^{\prime \prime }}(x) = - \dfrac{{\left[ {\left( {4{x^2} + 2x + 1} \right)(8) - (8x + 2)2 \times \left( {4{x^2} + 2x + 1} \right)(8x + 2)} \right]}}{{{{\left( {4{x^2} + 2x + 1} \right)}^4}}}\]
On substituting the value of \[x = - \dfrac{1}{4}\] in the above equation, we have
\[{f^{\prime \prime }}\left( { - \dfrac{1}{4}} \right) = - ve\]
Now, we know that \[{\rm{f}}({\rm{x}})\] is in maximum at
\[{\rm{x}} = - \dfrac{1}{4}\]
Now, we have determined the maximum value.
Maximum value is,
\[f{\left( { - \dfrac{1}{4}} \right)_{\max }} = \dfrac{1}{{4 \times \dfrac{1}{{16}} - 2 \times \dfrac{1}{4} + 1}} = \dfrac{1}{{\dfrac{1}{4} - \dfrac{2}{4} + 1}}\]
On simplifying the above equation, we obtain
\[\dfrac{4}{{1 - 2 + 4}}\]
Further simplification, we get
\[ = \dfrac{4}{3}\]
Therefore, if \[f(x) = \dfrac{1}{{\left[ {4{x^2} + 2x + 1} \right]}}\], then its maximum value is \[\dfrac{4}{3}\]

Therefore, the correct option is (A).

Note:The minimum and maximum values of a given function can also be determined using a different approach, called the differentiation method. We will calculate the value of the unknown after taking the first derivative, and then we will substitute that value in the second derivative of the function. If the value after substitution in the second derivative is negative, then the function has maximum value, and if it has positive value, then the derivative has minimum value.