Question

For all the real values of \[x\], the minimum value of \[\dfrac{{1 - x + {x^2}}}{{1 + x + {x^2}}}\]is
A. \[0\]
B. \[1\]
C. \[3\]
D. \[\dfrac{1}{3}\]
E. \[2\]

Answer 1

Hint: In this question, we have to find the minimum value of the given expression. The given expression is a fraction containing a linear equation in both of the numerator and the denominator. Also, the order of the linear equation on both of the sides of the fraction is two in each case. For doing this, we apply the first order derivative \[\left( {f'x} \right)\] on the given fraction. And since this is a fraction, we are going to have to apply the quotient rule on the fraction so as to find the first order derivative. Then, we simplify the fraction and put it equal to zero. Then we solve for the values of \[x\] and we are going to have our critical points for the expression. Then we just simply put in the critical values and the one giving off the minimum value is going to be our answer.

Formula Used:
We are going to use the formula of the first order derivative on the given expression which will be solved by applying the quotient rule and it is:
\[f'\left( {\dfrac{u}{v}} \right) = {\left( {\dfrac{u}{v}} \right)^\prime } = \dfrac{{u'v - v'u}}{{{v^2}}}\]

Complete step-by-step answer:
In the given question, the expression is \[f\left( x \right) = \dfrac{{1 - x + {x^2}}}{{1 + x + {x^2}}}\].
We need to find the minimum value of this expression.
For that, first we are going to find its first order derivative:
\[f'\left( x \right) = \dfrac{{d\left( {\dfrac{{1 - x + {x^2}}}{{1 + x + {x^2}}}} \right)}}{{dx}} = \dfrac{{{{\left( {1 - x + {x^2}} \right)}^\prime }.\left( {1 + x + {x^2}} \right) + {{\left( {1 + x + {x^2}} \right)}^\prime }.\left( {1 - x + {x^2}} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Solving the derivatives,
\[f'\left( x \right) = \dfrac{{\left( {2x - 1} \right).\left( {1 + x + {x^2}} \right) + \left( {2x + 1} \right).\left( {1 - x + {x^2}} \right)}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Multiplying the brackets,
\[f'\left( x \right) = \dfrac{{ - 1 - x - {x^2} + 2x + 2{x^2} + 2{x^3} - 1 + x - {x^2} - 2x + 2{x^2} - 2{x^3}}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Simplifying the numerator,
\[f'\left( x \right) = \dfrac{{2{x^2} - 2}}{{{{\left( {1 + x + {x^2}} \right)}^2}}}\]
Now, putting \[f'\left( x \right) = 0\],
\[\dfrac{{2{x^2} - 2}}{{{{\left( {1 + x + {x^2}} \right)}^2}}} = 0\]
Taking the denominator to the other side,
\[2{x^2} - 2 = 0\]
Taking 2 to the other side and taking the square root,
\[x = \pm 1\]
Hence, we have two critical points \[x = - 1, + 1\]
We are going to find the value of \[f\left( x \right)\] at these points and we have:
\[f\left( { - 1} \right) = \dfrac{{1 - \left( { - 1} \right) + {{\left( { - 1} \right)}^2}}}{{1 + \left( { - 1} \right) + {{\left( { - 1} \right)}^2}}} = 3\]
and \[f\left( 1 \right) = \dfrac{{1 - 1 + {1^2}}}{{1 + 1 + {1^2}}} = \dfrac{1}{3}\]

Hence, the correct option is D).

Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contains the known and the unknown and pick the one which is the most suitable for the answer. Then we put in the knowns into the formula, evaluate the answer and find the unknown.