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

Aydin Jarvis

Answered question

2022-10-13

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.

Answer & Explanation

espava8b

espava8b

Beginner2022-10-14Added 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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