Calculating an equation solution in C++

Question

I have a function in C++ with the following signature:

float Foo(float time, float min, float curr, float beta)

Within the function, I want to determine and return MAX in the following equation:

time = beta + (1.0f - beta) * ((MAX - curr) / (MAX - min))

To test the results, you could use the following arguments:

Foo(0.95f, 625, 800, 0.75f)

It should return 1500.

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On paper, I have the steps needed to determine MAX, but I don't know how to get that working in code. If anyone can provide the code to perform this calculation, I would be extremely grateful.

0.95 = 0.75 + (1 - 0.75) * ((max - 800) / (max - 625))
0.95 = 0.75 + 0.25 * ((max - 800) / (max - 625))

0.95 - 0.75 = 0.25 * ((max - 800) / (max - 625))
0.2 = 0.25 * ((max - 800) / (max - 625))

0.2 / 0.25 =  (max - 800) / (max - 625)
0.8 = (max - 800) / (max - 625)

0.8 * (max - 625) = max - 800
(0.8 * max) - (0.8 * 625) = max - 800
(0.8 * max) - 500 = max - 800

((0.8 * max) - max) - 500 = -800

((0.8 * max) - max) = -800 + 500
((0.8 * max) - max) = -300

-0.2 * max = -300

max = -300 / -0.2

max = 1500

Answer 1

time = beta + (1.0f - beta) * ((MAX - curr) / (MAX - min))

let's denote time with t, beta with b, curr with c, min with m, and MAX with x; we have

t = b + (1-b)(x-c)/(x-m)
(x-c)/(x-m) = (t-b)/(1-b)
(x-m)(t-b) = (x-c)(1-b)
x(t-b) - x(1-b) = m(t-b) - c(1-b)
x(t-1) = m(t-b) - c(1-b)
x = (m(t-b) - c(1-b))/(t-1)

so your function will be like this

float Foo(float time, float min, float curr, float beta)
{
   return (min*(time-beta) - curr*(1-beta))/(time-1);
}

I would also recommend refraining from using the identifiers min and time, because they may cause clashes with std::min and std::time

Answer 2

On the paper multiply each part of the equation on (MAX - min), then regroup elements to get MAX=some. The paper and the pencil is what you usually need before writing C++ functions like in your question.