Question

If $ f(x) = {x^2} - \dfrac{1}{{{x^2}}} $ then prove that $ f(x) = - f\left( {\dfrac{1}{x}} \right) $

Answer 1

Hint: First of all take the given expression and replace for the value of “x” here we will replace “x” with and then simplify the expression using the basic mathematical expression for the required resultant solution.

Complete step-by-step answer:
Take the given expression: $ f(x) = {x^2} - \dfrac{1}{{{x^2}}} $ …. (A)
Place the value $ x \to \dfrac{1}{x} $ in the above expression –
 $ f\left( {\dfrac{1}{x}} \right) = {\left( {\dfrac{1}{x}} \right)^2} - \dfrac{1}{{{{\left( {\dfrac{1}{x}} \right)}^2}}} $
When there is whole power then the power can be applied to the numerator as well as to the denominator.
 $ f\left( {\dfrac{1}{x}} \right) = \left( {\dfrac{{{1^2}}}{{{x^2}}}} \right) - \dfrac{1}{{\left( {\dfrac{{{1^2}}}{{{x^2}}}} \right)}} $
Square of the number can be given when the number is multiplied with itself. Here the square of one gives one only. Denominator’s denominator goes to the numerator.
 $ f\left( {\dfrac{1}{x}} \right) = \left( {\dfrac{1}{{{x^2}}}} \right) - \dfrac{{{x^2}}}{1} $
When there is number one in the denominator then the numerator can be expressed and written as only the term or as the numerator.
 $ f\left( {\dfrac{1}{x}} \right) = \left( {\dfrac{1}{{{x^2}}}} \right) - {x^2} $
Take minus one common on the right hand side of the equation and consider the concepts that minus minus gives plus and minus plus gives minus.
 $ f\left( {\dfrac{1}{x}} \right) = - \left[ { - \left( {\dfrac{1}{{{x^2}}}} \right) + {x^2}} \right] $
The above expression can be rearranged and re-written as –
 $ f\left( {\dfrac{1}{x}} \right) = - \left[ {{x^2} - \left( {\dfrac{1}{{{x^2}}}} \right)} \right] $
By using the value of the equation (A)
 $ f\left( {\dfrac{1}{x}} \right) = - \left[ {f\left( x \right)} \right] $
The above expression is the required solution.

Note: Be careful about the sign convention while simplifying the expression. Remember that the product of two negative terms gives the positive term and the product of one negative and the one positive term will give the negative term as the resultant value. Remember in any equation when you replace value for one term, the replacement should be done to the whole equation to get its equivalent value.