Find Min/Max of f: 2cos(x)+sin(2x) & x^2+16/x

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Discussion Overview

The discussion revolves around finding the absolute minimum and maximum values of two functions: \( f(x) = 2\cos(x) + \sin(2x) \) on the interval \([0, \frac{\pi}{2}]\) and \( f(x) = x^2 + \frac{16}{x} \) on the interval \([1, 3]\). Participants explore methods for determining extrema, including evaluating critical points and endpoints.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests that extrema occur where \( f'(x) = 0 \) and provides the derivative for the first function, leading to a quadratic equation in terms of \( \sin(x) \).
  • Another participant emphasizes the importance of checking both critical points and endpoints to determine global extrema on closed intervals.
  • A participant calculates the second derivative for the first function to confirm the nature of the critical point found.
  • For the second function, a participant finds the critical point by setting the derivative to zero and discusses the second derivative test to determine if it is a minimum.
  • There is a question about the correctness of the calculations and whether the identified points are indeed absolute extrema.

Areas of Agreement / Disagreement

Participants generally agree on the methods for finding extrema, but there is some uncertainty regarding the correctness of the calculations and whether all necessary evaluations have been made.

Contextual Notes

Participants note that global maxima and minima can occur at critical points or at the endpoints of the intervals, highlighting the need for comprehensive evaluation.

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4)Find the absolute minimum and absolute maximum values of f on the given interval.

a) f(x) = 2cos (x) + sin (2x) on [0, π/2]

b) f(x) = x^2 +16/x on [1, 3]
Thanks again in advance!
 
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What have you tried? What method do you know for finding extrema?
 
ayahouyee said:
4)Find the absolute minimum and absolute maximum values of f on the given interval.

a) f(x) = 2cos (x) + sin (2x) on [0, π/2]

b) f(x) = x^2 +16/x on [1, 3]
Thanks again in advance!

Hint: Global maxima and minima can occur either at turning points or endpoints.

How do you evaluate the turning points? How do you evaluate the endpoints?
 
I know that max/min are when f'(x)=0 so

a) f'(x)=-2sin(x)+2cos(2x) and
-2sin(x)+2cos(2x)=0
-sin(x)+cos(2x)=0 and cos(2x)=1-2sin^2(x)
so -sin(x)+1-2sin^2(x)=0
2sin^2(x)+sin(x)-1=0
so (2sinx-1)(sinx+1)=0
so sin(x)=1/2 or sin(x)=-1
giving x=π/6 on [0,π/2]
f"(x)=-2cos(x)-4sin(2x)
= -√3-2√3
f''(x) is <0 so f has a max at x=π/6 and it is an abs max there.

b)f'(x)=2x -16/x² =0 gives
x^3-8=0
gives x=2
f"(x)=2+32/x^3 =6 where x=2
Since f"(2)>0 x=2 is at a min point (2,12)
and on [1,3] this is an abs min pointis this correct? :)
 
Yes, except that the global maximum or minimum of a function on a closed segment can also lie on the boundary of the segment (see Wikipedia). So you should check the values of the functions at the ends of the intervals and choose the greatest or smallest between those and the values at critical points.
 

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