Prove that f ( x ) = 3 has a solution on the interval [ a , b ]And Intermediate Value Theorem says thatif f ( a ) ∗ f ( b ) &lt; 0, then it has a solution on that intervalSo instead of seeing if 3 is between the interval and stuff like that. Can't I just do this: f ( x ) = 3 f ( x ) − 3 = 0And then we consider f ( x ) − 3 a completely new whole function called g ( x ) = 0According to the theorem, I can say that since g ( a ) ∗ g ( b ) &lt; 0, it has a solution on the interval.

Question

Prove that   f  (  x  )  =  3 has a solution on the interval   [  a  ,  b  ]And Intermediate Value Theorem says thatif   f  (  a  )  ∗  f  (  b  )  &amp;lt;  0, then it has a solution on that intervalSo instead of seeing if 3 is between the interval and stuff like that. Can&#039;t I just do this:  f  (  x  )  =  3  f  (  x  )  −  3  =  0And then we consider   f  (  x  )  −  3 a completely new whole function called   g  (  x  )  =  0According to the theorem, I can say that since   g  (  a  )  ∗  g  (  b  )  &amp;lt;  0, it has a solution on the interval.

vard6vv · Accepted Answer

Yes, you can do this. There is a variant of this argument which allows you to prove the Brouwer fixed point theorem in one dimension: suppose   f  :  [  a  ,  b  ]  →  [  a  ,  b  ] is a continuous function. Show there exists a   c  ∈  [  a  ,  b  ] with   f  (  c  )  =  c.It&#039;s a great exercise to try out, and from your question you are halfway there already.

Prove that f ( x ) = 3 has a solution on the interval [ a , b ]

Answered question

Answer & Explanation

New Questions in Calculus 2