Question

OP - Operations with Polynomials Discussion

Let's say we are given three functions with the same domain.

a(x),
b(x), and
c(x), where c(x) = a(x) + b(x)
Each function has a maximum value (doesn't go to infinity, but rather stops at a maximum point).

S = the maximum value of a(x),
T = the maximum value of b(x), and
R = the maximum value of c(x).
Could there be a scenario where S + T = R for all cases? Is this always the case?

In this discussion, we will not fully answer the question (at first) but rather help each other come to a conclusion by posting an original argument of a case where S + T = R or a case where S + T ≠ R.

You must fully explain your case with equations or a picture using a graphing calculator (Links to an external site.).

Answer 1

Step-by-step 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.