If f (x-1) = 2x & # 178; + 1, find the analytic expression of F (x)

If f (x-1) = 2x & # 178; + 1, find the analytic expression of F (x)

Let t = X-1, then x = t + 1, so f (T) = 2 (T + 1) ^ 2 + 1 = 2T ^ 2 + 4T + 3, that is, f (x) = 2x ^ 2 + 4x + 3

Given the function f (x-1 / x) = x & # 178; + 1 / X & # 178;, then the analytic expression of function f (1-x) is

f(x-1/x)=x^2-2+1/x^2+2=(x-1/x)^2+2
So f (x) = x ^ 2 + 2
f(1-x)=(1-x)^2+2=x^2-2x+3

Given the function f (x-1) = x & # 178; - 2x-7, then f (4) =?

f(x-1)=x²-2x-7
f(x-1)=（x-1）²-8
f(4)=4²-8=8

For example, f (x + 4) = x2 + 6 to find f (x)
Use t = x + 4, x = T-4, and then substitute it into x2 + 6 to get f (T) = (T-4) 2 + 6 = t2-8t + 22, and then use X to replace T. I know why this method can replace t in the end. Is the replaced x equal to the previous x in F (x + 4)?

If I add two more steps, you should understand why
If you find the function expression of F (T), and then substitute t = x + 4, you will get:
f（x+4）=（x+4）^2-8(x+4)+22
This function expression means: the independent variable is x + 4, the dependent variable is f (x + 4), and its expression is "(x + 4) ^ 2-8 (x + 4) + 22";
So if the independent variable is x, should the dependent variable be f (x), and its expression be "x2-8x + 22"?
In a function, these x and T are not unknowns, but independent variables. They are a set of numbers. We can't use the concept of unknowns in the equation to look at them