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Question: If \[f\left( x \right) = \dfrac{{ax + b}}{{cx + d}}\], and \( fof\left( x \right) = x \) f or what g...

If f(x)=ax+bcx+df\left( x \right) = \dfrac{{ax + b}}{{cx + d}}, and fof(x)=xfof\left( x \right) = x f or what general condition.

Explanation

Solution

The term fof(x)fof\left( x \right) simply means composition of functions. It mathematically can be defined as applying one function to the results of another. So to find the general condition for the given question we first need to find fof(x)fof\left( x \right) . To find fof(x)fof\left( x \right) we need to replace the term xx in f(x)f\left( x \right) with f(x)f\left( x \right) .
So by using the above steps we can solve the given question for fof(x)fof\left( x \right) .

Complete step by step answer:
Given
Given

f(x)=ax+bcx+d  ......................(i) fof(x)=x..............................(iii)  f\left( x \right) = \dfrac{{ax + b}}{{cx + d}}\;......................\left( i \right) \\\ fof\left( x \right) = x..............................\left( {iii} \right) \\\

Now we have to find the general condition where fof(x)=xfof\left( x \right) = x
Now we know that mathematically fof(x)fof\left( x \right) simply means to replace the term xx in f(x)f\left( x \right) with f(x)f\left( x \right) .
So we can write:
fof(x)=a(ax+bcx+d)+bc(ax+bcx+d)+d  ......................(iii)fof\left( x \right) = \dfrac{{a\left( {\dfrac{{ax + b}}{{cx + d}}} \right) + b}}{{c\left( {\dfrac{{ax + b}}{{cx + d}}} \right) + d}}\;......................\left( {iii} \right)
Now let’s simplify the equation (iii) since we can then apply the given condition.
So we can write:

fof(x)=a(ax+bcx+d)+bc(ax+bcx+d)+d   fof(x)=a(ax+b)+b(cx+d)c(ax+b)+d(cx+d) fof(x)=a2x+bx+bcx+bdcax+cb+dcx+d2....................(iv)     fof\left( x \right) = \dfrac{{a\left( {\dfrac{{ax + b}}{{cx + d}}} \right) + b}}{{c\left( {\dfrac{{ax + b}}{{cx + d}}} \right) + d}}\; \\\ fof\left( x \right) = \dfrac{{a\left( {ax + b} \right) + b\left( {cx + d} \right)}}{{c\left( {ax + b} \right) + d\left( {cx + d} \right)}} \\\ fof\left( x \right) = \dfrac{{{a^2}x + bx + bcx + bd}}{{cax + cb + dcx + {d^2}}}....................\left( {iv} \right) \\\ \; \\\

Now we have simplified the expression of fof(x)fof\left( x \right) . So after simplifying it we have to equate our equation with the given condition.
It has been given that fof(x)=xfof\left( x \right) = x , so on equation it to our equation we can write:
fof(x)=a2x+bx+bcx+bdcax+cb+dcx+d2=x....................(v)fof\left( x \right) = \dfrac{{{a^2}x + bx + bcx + bd}}{{cax + cb + dcx + {d^2}}} = x....................\left( v \right)
Now let’s solve and simplify the equation (v) such that we can find the general condition for the given question.
So we get:

fof(x)=a2x+bx+bcx+bdcax+cb+dcx+d2=x a2x+bx+bcx+bd=x(cax+cb+dcx+d2) x(a2+b+bc+bdx)=x(cax+cb+dcx+d2) a2+b+bc+bdx=cax+cb+dcx+d2..................(vi)  fof\left( x \right) = \dfrac{{{a^2}x + bx + bcx + bd}}{{cax + cb + dcx + {d^2}}} = x \\\ {a^2}x + bx + bcx + bd = x\left( {cax + cb + dcx + {d^2}} \right) \\\ x\left( {{a^2} + b + bc + \dfrac{{bd}}{x}} \right) = x\left( {cax + cb + dcx + {d^2}} \right) \\\ {a^2} + b + bc + \dfrac{{bd}}{x} = cax + cb + dcx + {d^2}..................\left( {vi} \right) \\\

Now we can see that to satisfy the LHS and RHS of the equation (vi) the only condition possible is a=−da = - d .Therefore our final answer is a=−da = - d .

Note: The notation "f (x)" is exactly the same thing as "y". The expression "f (x)" means "a formula, named f, has x as its input variable" rather than “multiply f and x". Also another thing one should always remember is that the function name is what comes before the parentheses.
Also similar questions to find the term fof(x)fof\left( x \right) or to find composition of functions one should follow the above described steps which are simple and efficient.