find my algebraic error please?

f(x) = (x^2 + 400)^(.5) + 60 + (1600 + (200 - x)^2)^.5

Let's clean everything up in the second set of brackets first:

f(x) = (x^2 + 400)^.5 + 60 + (1600 + 40000 - 400x + x^2)^.5

f(x) = (x^2 + 400)^(.5) + (x^2 - 400x + 41600)^(.5) + 60

f'(x) = (2x) * (.5) / sqrt(x^2 + 400) + (2x - 400) * (.5) / sqrt(x^2 - 400x + 41600)

f'(x) = x / sqrt(x^2 + 400) + (x - 200) / sqrt(x^2 - 400x + 41600)

Get a common denominator:

f'(x) = x * sqrt(x^2 - 400x + 41600) + (x - 200) * sqrt(x^2 + 400) / [sqrt(x^2 + 400) * (sqrt(x^2 - 400x + 41600)]

We can disregard the denominator since we want to know when f'(x) = 0

0 = x * sqrt(x^2 - 400x + 41600) + (x - 200) * sqrt(x^2 + 400)

-x * sqrt(x^2 - 400x + 41600) = (x - 200) * sqrt(x^2 + 400)

x^2 * (x^2 - 400x + 41600) = (x^2 - 400x + 40000) * (x^2 + 400)

x^4 - 400x^3 + 41600x^2 = x^4 + 400x^2 - 400x^3 - 160000x + 40000x^2 + 16000000

x^4 - x^4 - 400x^3 + 400x^3 + 41600x^2 - 40400x^2 + 160000x - 16000000 = 0

1200x^2 + 160000x - 16000000 = 0

12x^2 + 1600x - 160000 = 0

3x^2 + 400x - 40000 = 0

a = 3

b = 400

c = -40000

x = (-400 +/- sqrt(160000 + 12 * 40000)) / 6

x = (-400 +/- sqrt(160000 + 480000)) / 6

x = (-400 +/- sqrt(640000)) / 6

x = (-400 +/- 800) / 6

x = 400 / 6

x = -1200 / 6

We want positive x

x = 200 / 3

it's normal, your function is strictly increasing in the range 0-200, even in the range of all positive values

(you are adding positive function that grow with x when x is positive).The maximum value in your range is at 200 (the maximum value of x)