Putnam Exam Challenge (Maximum Value)

  • Thread starter Amad27
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  • #26
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Simply because f(x) = g(u).

Hi,

Ok, so let us summarize.

(1)The domain of g(u) is the range of u = sin(x) + cos(x) which is [itex]-√2 ≤ u ≤ √2[/itex]
(2) The range of f(x) is the range of g(u)

You still cant treat "u" as an independent variable, you must apply the chain rule. Just as you cant do this.

Find the derivative of [itex]f(x) = (2x^2 + 3x)^2[/itex]

Suppose you let u = 2x^2 + 3x then

[itex]f(x) = u^2 → f'(x) = 2u = 4x^2 + 6x[/itex] wrong
 
  • #27
disregardthat
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Amad, that really does not matter. We are not assuming or requiring that the derivatives are equal.
 
  • #28
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Amad, that really does not matter. We are not assuming or requiring that the derivatives are equal.

Hello,

Out of curiosity, I graphed both,

[itex]f_1(x) = sin(x) + cos(x) + tan(x) + sec(x) + csc(x)[/itex] and
[itex]g(u) = u + 2/(u-1)[/itex]

The interval (in terms of u was) [itex]-√2 ≤ u ≤ √2[/itex]

They do not overlap what soever, the minimum of

g(x) = x + 2/(x-1) is NOT the range of f_1(x) = sin(x) + cos(x) + tan(x) + cot(x) + sec(x) + csc(x)

Check out this approach.

Let [itex]h(x) = u + 2/(u-1) & u = sin(x) + cos(x) → -√2 ≤ u ≤ √2[/itex]
[itex] let v = cos(x) - sin(x)[/itex]

[itex] h'(x) = v + 2[-v]/[[u-1]^2 [/itex]
[itex]h'(x) = v - 2v/[u - 1]^2[/itex]

Set h'(x) = 0

[itex]h'(x) = 0 → v - 2v/[u-1]^2 = 0 → v[1 - 2/[u-1]^2] = 0[/itex]
[itex]v = 0 → x = pi/4 [/itex]
[itex] (u-1)^2 = 2 → u = √2 + 1 → u = -√2 + 1[/itex]

We know that

[itex] h(x) = f(u)[/itex] since this is a simple rearranging, and grouping...

h(x) at the corresponding "x" when u = √2 + 1 for example is f(√2 + 1)

This just seems more correct because it is just unusual treating a composite function u = cos(x) + sin(x) as an independent variable like u = x.
 

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