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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.

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  • $\begingroup$ Hint (for $a$): Let $f(x)=e^x+x^2-2$. What is $f(0)$? What is $f(1)$? For $b$..what does "sen" mean? $\endgroup$
    – lulu
    Commented Jul 10, 2017 at 0:21
  • $\begingroup$ What is $sen(x)$? $\endgroup$
    – cat
    Commented Jul 10, 2017 at 0:21
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    $\begingroup$ "Their graphs have 2 long vertical lines with medium space between them"... I do not know what you tried to graph. Did you try to graph $e^x=2-x^2$... which has nothing to do with $y$ at all? You should have graphed the one function $y=e^x$ and graph the other function $y=2-x^2$ on top of that one to see the intersection like so. $\endgroup$
    – JMoravitz
    Commented Jul 10, 2017 at 0:22
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    $\begingroup$ @cat The asker is likely from a Spanish- or Portuguese-speaking country. $\mathrm{sen}(x)\equiv \sin(x)$ $\endgroup$
    – Michael L.
    Commented Jul 10, 2017 at 0:23
  • $\begingroup$ @MichaelLee thanks, I already fixed $\endgroup$
    – Goun2
    Commented Jul 10, 2017 at 0:26

2 Answers 2

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hint for the first

Let $f (x)=e^x-2+x^2$.

  • $f $ is continuous at $[0,1] $

  • $f (0)=-1<0$

  • $f (1)=e-1>0$

  • $0\in (f (0),f (1)) $

hence there exist $c\in (0,1) $ such that $f (c)=0$ or $$e^c=2-c^2$$

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  • $\begingroup$ Thanks for the help, is this e the euler's number ?, and to solve the second one I should use the values in radians ? $\endgroup$
    – Goun2
    Commented Jul 10, 2017 at 0:41
  • $\begingroup$ @hjx You can take zero and pi. $\endgroup$ Commented Jul 10, 2017 at 0:46
  • $\begingroup$ Clear up something, Why did you passed e^k to the right hand side as positive and 2 got negative ? $\endgroup$
    – Goun2
    Commented Jul 10, 2017 at 2:02
  • $\begingroup$ @hjx note: $e^x=2-x^2$ iff $e^x\color{red}{+x^2-2}=2-x^2\color{red}{+x^2-2}$ iff $e^x+x^2-2=0$. We can add and subtract the same thing to both sides without changing the truth of the statement $\endgroup$
    – JMoravitz
    Commented Jul 10, 2017 at 2:36
  • $\begingroup$ @JMoravitz thanks one more time,You are the man. $\endgroup$
    – Goun2
    Commented Jul 10, 2017 at 2:52
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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).

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  • $\begingroup$ thanks for the help. $\endgroup$
    – Goun2
    Commented Jul 10, 2017 at 0:52

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