4(0.5n-3)=n-0.25(12-8n)
2n-12=n-3+2n
2n-12=3n-3
2n-3n=-3+12
-n=9
n = -9
Assuming we need to find i such that
1 ≤ i ≤ n and t[i]=i.
If we need to find only the first occurrence, we can do:
for i:1 to n {
if t[i]=i then return(i)
}
If exhaustive search is required, then put the results (values of i) in an array or a linked list, return the number of values found, and the array (or linked list).
if indeed two functions are inverse of each other, then their composite will render a result of "x", namely, if g(x) is indeed an inverse of f(x), then
![\bf (g\circ f)(x)=x\implies g(~~f(x)~~)=x \\\\\\ \begin{cases} f(x) = 3x\\ g(x)=\cfrac{1}{3}x \end{cases}\qquad \qquad g(~~f(x)~~)=\cfrac{1}{3}[f(x)]\implies g(~~f(x)~~)=\cfrac{1}{3}(3x)](https://tex.z-dn.net/?f=%5Cbf%20%28g%5Ccirc%20f%29%28x%29%3Dx%5Cimplies%20g%28~~f%28x%29~~%29%3Dx%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20f%28x%29%20%3D%203x%5C%5C%20g%28x%29%3D%5Ccfrac%7B1%7D%7B3%7Dx%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20g%28~~f%28x%29~~%29%3D%5Ccfrac%7B1%7D%7B3%7D%5Bf%28x%29%5D%5Cimplies%20g%28~~f%28x%29~~%29%3D%5Ccfrac%7B1%7D%7B3%7D%283x%29)