A Find the minimum without Calculus or Graphing

AI Thread Summary
The discussion focuses on finding the minimum of the function ƒ(ß)=0.5sec(ß) + √[1+(sec²(ß)/4)+tan(ß)/√(2)] without using calculus or graphing. The original poster already knows the minimum occurs at π/12 and is approximately 1.5 but seeks a method to derive this result. Suggestions include using trigonometric identities for simplification and creating a table of values to identify the lowest points. Additionally, fitting the three lowest adjacent values to a quadratic can help find the minimum using the standard formula for a parabola. The conversation emphasizes exploring analytical methods rather than computational tools.
patrickbotros
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Poster is on a 10-day vacation from the PF for repeatedly misplacing schoolwork type questions
ƒ(ß)=.5sec(ß) + √[1+(sec2(ß)/4)+tan(ß)/√(2)]
Without graphing it or using calculus find the minimum. I already know the answer but want to know how to do it. It s at π/12 and is something like 1.5.
First off this is NOT a homework problem. I already know the answer is something like 1.5 at π/12 but I want to know how to do it. DONT PUT THIS IN A HOMEWORK SECTION.
 
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Perhaps one can use trig identities to simplify it first.
 
I'd like to know the actual function. Probably not
patrickbotros said:
ƒ(ß)=.5sec(ß) + √[1+(sec2(ß)/4)+tan(ß)/√(2)]
or, as we write it here $$ {1\over 2\cos\beta}+ \sqrt{ 1 + {1\over 4 \cos^2\beta} + {1\over \sqrt 2}\tan\beta} $$ surely ?
 
Probably not.
f(Pi/12)=1.73
f(0)=1.62

@OP
Have you checked if the Pi/12 and 1.5 satisfy the expression?
 
Make a large table of values at small step size.

Pick the three lowest adjacent values.

Fit to a quadratic.

Use the standard formula for the minimum of a parabola opening upward.
 
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