[ I WILL GIVE BRAINLIEST ] to the person who can explain as to how to do an equation like this. I need an explanation because I want to understand the concept. How do you do an equation like this? ( LOOK AT IMAGE PLEASE )

Answer:
f(g(x)) = x
Step-by-step explanation:
A function is a relation that maps an input to an output. The inverse of that function is the relation that takes that output value and maps it to the input that produced it. The range of a function is the domain of its inverse, and vice versa.
function: y = f(x)
inverse function: x = f⁻¹(y)
We have defined the inverse function in this problem to be ...
g(x) = f⁻¹(x)
The result of a function operating on its inverse (and vice versa) is the original value, unchanged:
f(g(x)) = x
g(f(x)) = x
When a function and its inverse are graphed, the graphs are mirror images of each other across the line y=x.
To show functions f and g are mutual inverses algebraically, we can demonstrate that ...
f(g(x)) = x
Simplifying this expression, we have ...
[tex]f(g(x))=\dfrac{g(x)+5}{2g(x)+1}=\dfrac{\dfrac{5-x}{2x-1}+5}{2\dfrac{5-x}{2x-1}+1}=\dfrac{(5-x)+5(2x-1)}{2(5-x)+(2x-1)}\\\\=\dfrac{5-x+10x-5}{10-2x+2x-1}=\dfrac{9x}{9}\\\\=\boxed{f(g(x))=x}[/tex]
The attached graph shows the functions are reflections of each other across the line y=x.
We can also find the inverse of the function f(x) and show it is equal to g(x).
[tex]x = f(y)\\\\x=\dfrac{y+5}{2y+1}\qquad\text{elaborate $f(x)$}\\\\x(2y+1)=y+5\qquad\text{multiply by $(2y+1)$}\\\\2yx-y=5-x\qquad\text{add $(-x-y)$}\\\\y(2x-1)=5-x\qquad\text{factor out $y$}\\\\y=\dfrac{5-x}{2x-1}=g(x)\qquad\text{divide by $y$-coefficient, compare to $g(x)$}[/tex]
Additional comment
You may recall that addition of a value and its additive inverse results in the additive identity element (0). (x) +(-x) = 0
Similarly, multiplication of a value by its multiplicative inverse results in the multiplicative identity element (1). (x) × (1/x) = 1
The general case of a function operating on its inverse resulting in identity is used at many levels in math. In this problem, we're introduced to the fact that a function operating on its inverse leaves the original input unchanged. In matrix algebra, we see that the product of a matrix and its inverse is the identity matrix.
In algebra, there are a few function/inverse-function pairs that we regularly use: powers and roots, logarithms and exponential functions, trig functions and their inverses. The handheld calculators we have access to have calculation of these pairs built in.