a) e x = 2 − x 2 b) sin ⁡ ( x ) = x 2 − 1The 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.

Question

a)          e    x    =  2  −      x    2  b)      sin  ⁡  (  x  )  =      x    2    −  1The 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&#039;t return exact numbers, I&#039;m confused, if anyone can give more clarifications about these problems it would be of great help for me.

Jake Landry · Accepted Answer

hint for the firstLet   f  (  x  )  =      e    x    −  2  +      x    2  .1.   f is continuous at [0,1]2.   f  (  0  )  =  −  1  &amp;lt;  03.   f  (  1  )  =  e  −  1  &amp;gt;  04.   0  ∈  (  f  (  0  )  ,  f  (  1  )  )hence there exist   c  ∈  (  0  ,  1  ) such that   f  (  c  )  =  0 or      e    c    =  2  −      c    2

imire37 · Accepted Answer

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).

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

Answered question

Answer & Explanation

New Questions in Calculus 2