Question

If $f(x)=ax+b$ and $g(x)=cx+d$, then $f({f(x)})=g({f(x)})$
 is equivalent to?
A. $f(a)=g(c)$
B. $f(b)=g(b)$
C. $f(d)=g(b)$
D. $f(c)=g(a)$

Answer 1

Hint: If two functions $f$ and $g$ have the same domain and codomain, and if $f(a)=g(a)$ for every a in the domain, then we say that they are equal. Here we will use the composition of functions. By using the individual functions as a reference, replace the variable $x$ that is present in the external function with the internal function. Finally, simplify the function that was obtained.

Formula Used: The composition of functions is given by:
$fog(x)= f[g(x)]$
$gof(x)= g[f(x)]$
Here $f(x)$ and $g(x)$ are two functions.

Complete step by step solution: Given functions are, $f (x) = ax + b$ and $ g(x) = cx + d$.
Since, these function are equivalent
$\therefore f [g(x)] = g[f (x)] \\
\Rightarrow f(cx + d) = g(ax + b)\\
\Rightarrow a(cx + d) + b = c (ax + b) + d\\
\Rightarrow acx + ad + b = acx + bc + d\\
\Rightarrow ad + b = cb + d\\
f(d) = g(b)$

So, option C is correct.

Note: The process of merging two or more functions into one function is called the composition of functions. Remember that each statement for the composition uses the same letters in the same order.
You should begin with function g since $f (g(x))$ makes this clear (innermost parentheses are done first).
It should be noticed that if the range of $f$ is a subset of $g$, then $gof$ exists. In the same way, if the range of $g$ is a subset of domain $f$, then $fog$ exists.