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Question: If for non-zero \(x,af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5,\)where \(a \ne b,\) ...

If for non-zero x,af(x)+bf(1x)=1x−5,x,af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5,where a≠b,a \ne b, then find f(x)f(x).

Explanation

Solution

AS we know that the above question consists of a functional equation. A functional equation is any equation in which the unknown represents the function. We know that this type of function assigns exactly one output to each specified type. It is common to name the functions f(x)f(x) or g(x)g(x). These functional equations have a common technique for solving the value of f(x)f(x). We will first replace xx by 1x\dfrac{1}{x}.

Complete step by step solution:
As per the given question we have af(x)+bf(1x)=1x−5af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5 and we have to find the value of f(x)f(x).
We will first replace the xx by 1x\dfrac{1}{x} and we have: af(1x)+bf(x)=x−5af\left( {\dfrac{1}{x}} \right) + bf\left( x \right) = x - 5. We will now multiply the first equation with aa and the second equation by bb, then we have a2f(x)+abf(1x)=ax−5a{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) = \dfrac{a}{x} - 5a (we call it the third equation), now the another equation is abf(1x)+b2f(x)=bx−5babf\left( {\dfrac{1}{x}} \right) + {b^2}f(x) = \dfrac{b}{x} - 5b(it is the fourth equation). After this we will subtract the equation fourth from the third equation, we have: a2f(x)+abf(1x)−abf(1x)−b2f(x)=ax−5a−(bx−5b){a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right) - {b^2}f(x) = \dfrac{a}{x} - 5a - (\dfrac{b}{x} - 5b), solving this we get a2f(x)−b2f(x)=ax−5a−bx+5b{a^2}f(x) - {b^2}f(x) = \dfrac{a}{x} - 5a - \dfrac{b}{x} + 5b. In this equation we will take the common factor out and we have, (a2−b2)f(x)=(a−b)1x−5a+5b\left( {{a^2} - {b^2}} \right)f(x) = (a - b)\dfrac{1}{x} - 5a + 5b. We know the algebraic
formula that (a2−b2)=(a+b)(a−b)({a^2} - {b^2}) = (a + b)(a - b). So we can substitute this and we get,(a−b)(a+b)f(x)=(a−b)1x−5(a−b)(a - b)(a + b)f(x) = (a - b)\dfrac{1}{x} - 5(a - b) ⇒(a+b)f(x)=1−5xx \Rightarrow (a + b)f(x) = \dfrac{{1 - 5x}}{x} .
Now by isolating the term we get f(x)=(1−5x)x(a+b)f(x) = \dfrac{{(1 - 5x)}}{{x(a + b)}}.
Hence the required answer is f(x)=(1−5x)x(a+b)f(x) = \dfrac{{(1 - 5x)}}{{x(a + b)}}.

Note: Before solving this type of question we should the function equation, their formulas and method to solve it. We should also have the knowledge of the algebraic identities as they are very useful in calculation of this kind of problem. We should note that in 1x\dfrac{1}{x}, xx is replaced also which can be written as 11x=x1\dfrac{1}{{\dfrac{1}{x}}} = \dfrac{x}{1}.