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Let a function of domain equal to $\mathbb{R}$ be $f(x)=e^x-3$.

In which of the follows intervals, by the Bolzano theorem, we can say that $f(x)=-x-\frac{3}{2}$ have at least one solution?

$A) \left ]0,\frac{1}{5} \right[$

$B) \left ]\frac{1}{5},\frac{1}{4} \right[$

$C) \left ]\frac{1}{4},\frac{1}{3} \right[$

$D) \left ]\frac{1}{3},1 \right[$

At first I tried to solve $\space e^x-3=-x-\frac{3}{2}\space$ in order of $x$. I riched the equation $\space e^x+x=\frac{3}{2} \space$that I don't know how to solve, with the tools that I have learned.

I know that the$\space e^x-3=-x-\frac{3}{2}\space$ solution's is the intersept point beteewn the two functions. But I can't manage the solution using the Bolzano theorem principles.

Thanks for the help

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Try writing your equation in the form $g(x)=0$ for some function $g$ (this function will involve only $e^x,x$ and a constant number). Can you verify that $g$ is a continuous function? What does Bolzano's Theorem tell you about the behaviour of a continuous function $g$ on an interval $[a,b]$? –  user12477 Jun 22 '12 at 22:48

3 Answers 3

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You don't have to solve the equation, because you only want to know if there is a solution. Remember that Bolzano's theorem says that if $f$ is a continuous function in $[a,b]$, and the signs of $f(a)$ and $f(b)$ are different (one positive and the other negative), then there is at least one root of $f$ in $(a,b)$, i.e. a value of $x \in (a,b)$ with $f(x) = 0$.

So what you want to do is first come up with a function that will work: since $e^x - 3 = - x - \frac{3}{2} \Leftrightarrow e^x - 3 + x + \frac{3}{2} = 0$, you can pick $f(x) = e^x - 3 + x + \frac{3}{2}$. Now for each one of the intervals, calculate the images of the endpoints, and if the signs are different then there is a root of $f$ in that interval.

Note that there could be a root in those intervals even if the images of the endpoints have different signs, but we can't know by Bolzano's theorem.

  1. $f(0) \lt 0$ and $f\left(\frac{1}{5}\right) \lt 0$, so we can't assure by Bolzano's theorem that there's a root in $\left(0,\frac{1}{5}\right)$.
  2. $f\left(\frac{1}{5}\right) \lt 0$ and $f\left(\frac{1}{4}\right) \gt 0$, so Bolzano's theorem tells us that there's at least one root in $\left(\frac{1}{5},\frac{1}{4}\right)$.
  3. $f\left(\frac{1}{4}\right) \gt 0$ and $f\left(\frac{1}{3}\right) \gt 0$, so again we can't be sure if there's a root in $\left(\frac{1}{4},\frac{1}{3}\right)$.
  4. $f\left(\frac{1}{3}\right) \gt 0$ and $f(1) \gt 0$, so we can't be sure if there's a root in $\left(\frac{1}{4},1\right)$.
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I didn't even thought to pass all terms to the same side of the equation, using the rules of functions operations. I chosen this answer, because is the most complete.Thanks to everyone –  João Jun 22 '12 at 23:20
    
@João Bolzano's theorem applies to a function $f$ and gives sufficient conditions so that $f(x)=0$ for some $x$ in an interval. You need to find a function first, by passing all the terms to one side of the equation, so that you have $0$ on the other side. –  talmid Jun 22 '12 at 23:23

Let $h(x)=e^x-3-\left(-x-\frac32\right)=e^x+x-\frac32$. Bolzano’s theorem, or the intermediate value theorem, says that if $h(x)$ is positive at one endpoint of an interval and negative at the other, it must be $0$ somewhere in the interval. Thus, you need only check the signs of $h(0),h(1/5),h(1/4),h(1/3)$, and $h(1)$ to see what the theorem tells you about the existence of solutions to $f(x)=-x-\frac32$ in the intervals that you’ve been given.

Added: As a hint, there is exactly one interval in which $f(x)=-x-\frac32$ has a solution.

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Let $\,g(x)=e^x-3+x+\frac{3}{2}\,$ . Now just check that $$g\left(\frac{1}{5}\right)<0\,\,,\,\,g\left(\frac{1}{4}\right)>0$$ and apply Bolzano's theorem...

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