a) e^x=2−x^2 b) sin(x)=x^2−1 use the intermediate value theorem to prove the following equations have solution.

popljuvao69

popljuvao69

Answered question

2022-08-11

a)   e x = 2 x 2
b)   sin ( x ) = x 2 1
The intermediate value theorem says if F is a continues function in a closed interval [ A , B ], and you choose a value K between f ( A ) and f ( B ) where f ( A ) f ( B ), Then, there is a value between A and B, C, such that f ( C ) = K.

I tried to solve the first one by trying to do what the intermediate value theorem says. Then I went on by choosing 2 values A and B. But the first equation has a weird behavior and so does the second one.

Their graphs have 2 long vertical lines with a medium space between them, and Choosing values to plug in doesn't return exact numbers, I'm confused, if anyone can give more clarifications about these problems it would be of great help for me.

Answer & Explanation

Jake Landry

Jake Landry

Beginner2022-08-12Added 22 answers

hint for the first
Let f ( x ) = e x 2 + x 2 .

1. f is continuous at [0,1]
2. f ( 0 ) = 1 < 0
3. f ( 1 ) = e 1 > 0
4. 0 ( f ( 0 ) , f ( 1 ) )

hence there exist c ( 0 , 1 ) such that f ( c ) = 0 or
e c = 2 c 2
imire37

imire37

Beginner2022-08-13Added 8 answers

Showing that e x = 2 x 2 has a solution is the same as showing the function f ( x ) = e x 2 + x 2 has a zero. Use the intermediate value theorem on f ( x ) (note f ( x ) is continuous).

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?