I'm taking differential calculus, and I've read that is actually a pretty interesting function that just has one solution. I'm a little confused since most proofs that involve the Intermediate value theorem give a closed interval. But I need to prove that it has a solution in the real numbers. Thanks in advance.

Aydin Jarvis 2022-10-13 Answered
I'm taking differential calculus, and I've read that is actually a pretty interesting function that just has one solution. I'm a little confused since most proofs that involve the Intermediate value theorem give a closed interval. But I need to prove that it has a solution in the real numbers. Thanks in advance.
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Answers (1)

espava8b
Answered 2022-10-14 Author has 12 answers
HINT:
Let f ( x ) = x cos ( x ).
Note that f is continuous for x [ 0 , π / 2 ] with f ( 0 ) = 1 and f ( π / 2 ) = π / 2.
Given the Intermediate Value Theorem, what can one say about f on the closed interval [ 0 , π / 2 ], which is a subset of the real numbers?
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How can I prove my thoughts in a formal correct way? (Are my thoughts generally correct?)

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