Explanation:
S + T = R
Suppose we define ...
a(x) = 2x, for 0 ≤ x ≤ 1
b(x) = x^2, for 0 ≤ x ≤ 1
Then we have the following:
c(x) = a(x) +b(x) = 2x +x^2, for 0 ≤ x ≤ 1
S = max(a(x)) = a(1) = 2
T = max(b(x)) = b(1) = 1
R = max(c(x)) = c(1) = 2 +1 = 3
This value of R satisfies S + T = R.
We note that for x=p=1, we have S = a(p), T = b(p), and R = c(p). The first attachment illustrates this case.
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S + T ≠ R
Suppose we define ...
a(x) = x, for 0 ≤ x ≤ 1
b(x) = 1 -x^2, for 0 ≤ x ≤ 1
c(x) = a(x) +b(x) = x + 1 -x^2, for 0 ≤ x ≤ 1
Then we have the following:
S = max(a(x)) = a(1) = 1
T = max(b(x)) = b(0) = 1
R = max(c(x)) = c(0.5) = 1.25 ≠ 1 + 1 = 2
This value of R does not satisfy S + T = R.
We note that for p, q, r we have S = a(p), T = b(q), R = c(r) and p≠q≠r. The second attachment illustrates this case.
