Maxima minina on an interval (calculus+trig)

  • Thread starter Thread starter quicksilver123
  • Start date Start date
  • Tags Tags
    Interval Maxima
Click For Summary
SUMMARY

The discussion focuses on finding the maximum and minimum values of the function f(x) = x + cos(x) over the interval <-π, 2π>. The derivative of the function is correctly identified as f'(x) = 1 - sin(x). The critical insight is that f'(x) is non-negative since sin(x) is always less than or equal to 1 for all real x, indicating that the function is increasing on the specified interval. Understanding this derivative behavior is crucial for determining the extrema of the function algebraically.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Knowledge of trigonometric functions, particularly sine and cosine
  • Familiarity with interval notation and its implications in calculus
  • Ability to analyze function behavior without graphical representation
NEXT STEPS
  • Study the implications of critical points in calculus
  • Learn about the Mean Value Theorem and its applications
  • Explore the behavior of trigonometric functions within specified intervals
  • Investigate the concept of increasing and decreasing functions
USEFUL FOR

Students and educators in calculus, mathematicians interested in function analysis, and anyone looking to deepen their understanding of derivatives and extrema in trigonometric contexts.

quicksilver123
Messages
173
Reaction score
0
IMG_2066.jpg
I need to find the max/min of a function on an interval.
The function is f(x)=x+cos(x) and the interval is <-PI,2pi>
There is an attached solution but I do not understand how to arrive at the given solution (see screenshot). I would personally just take the derivative as
F'(x)=1-sin(x)
However the solution says this is only half the answer and I do not understand the reasoning (I am trying to do this algebraicly, without thinking of it graphically).
 
Physics news on Phys.org
quicksilver123 said:
the solution says this is only half the answer
I think you may be misinterpreting what it says. It is strangely worded. Better would be "f'(x)=1-sin(x), which is ≥0 since sin(x)≤1 for any real x".
 
  • Like
Likes   Reactions: quicksilver123
Thanks
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
10
Views
1K
Replies
32
Views
3K
  • · Replies 1 ·
Replies
1
Views
849
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
1K