Question

If $f(x) = \dfrac{{x + 2}}{{(3x - 1)}}$, then f\left\\{ {f(x)} \right\\} is equal to
A) $x$
B) $- x$
C) $\dfrac{1}{x}$
D) $- \dfrac{1}{x}$

Answer 1

In this question we have been given the value of $f(x) = \dfrac{{x + 2}}{{(3x - 1)}}$. And we have to find the value of f\left\\{ {f(x)} \right\\}.
We will put the value:
$\dfrac{{x + 2}}{{3x - 1}}$, in the place of $x$ in both numerator and denominator and then we solve the value.

Complete step by step answer:
We have been given that $f(x) = \dfrac{{x + 2}}{{(3x - 1)}}$.
We have to calculate f\left\\{ {f(x)} \right\\} . So by putting the value of $f(x)$ we can write the expression as $f(f(x) = \dfrac{{f(x) + 2}}{{f(x) - 1}}$
Now we simplify this:
$\dfrac{{\dfrac{{x + 2}}{{3x - 1}} + 2}}{{3\left( {\dfrac{{x + 2}}{{3x - 1}}} \right) - 1}}$
We will take the LCM in both numerator and denominator and it gives us:
$\dfrac{{\dfrac{{x + 2 + 2(3x - 1)}}{{3x - 1}}}}{{\dfrac{{3x + 6}}{{3x - 1}} - 1}} = \dfrac{{\dfrac{{x + 2 + 2(3x - 1)}}{{3x - 1}}}}{{\dfrac{{3x + 6 - 1(3x - 1)}}{{3x - 1}}}}$
Now by multiplying the values in the denominator we have $\dfrac{{\dfrac{{x + 2 + 6x - 2}}{{3x - 1}}}}{{\dfrac{{3x + 6 - 3x + 1}}{{3x - 1}}}}$
We can write the above expression also as
$\dfrac{{x + 2 + 6x - 2}}{{3x + 6 - 3x + 1}}$
By adding the similar terms together we have
$\dfrac{{7x}}{7} = x$
Hence, the correct option is (A) $x$.

Note:
We should note 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 $f(x)$ or $g(x)$.
These functional equations have a common technique for solving the value of $f(x)$.
A function $f(x)$ is known as continuous function.