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I am supposed to show that equation $ x^3=\ln(x+3)-2$ has at least one solution on an open interval $(-3,\infty)$ using mean value theorem. I know that this theorem needs to be used on close interval, so I don't know how to proceed.

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    $\begingroup$ You make the interval smaller and closed? $\endgroup$ – user370967 Jan 31 '18 at 12:04
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The problem boils down to determining the zeros of $f(x) = x^3+2-\ln(x+3)$ since $x=-3$ is not part of the domain you must have an open interval. If you plug in $x=-2$ and $x=0$ you can see the function changes sign and since it is continuous on its domain you must have $$ f([a,b]) = \left[\min_{[a,b]}f, \max_{[a,b]} f \right]$$ where $[a,b]$ inside the domain of $f(x)$. So at some $c \in [-2,0]$ you have $f(c) = 0$.

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Hint:

  • Using $x=-2$ gives $$x^3=(-2)^3=-8\color{red}<-2=\ln(1)-2=\ln(x+3)-2.$$
  • Using $x=e^2-3$ gives $$x^3=(e^2-3)^3>(2^2-3)^3=1\color{red}>0=\ln(e^2)-2=\ln(x+3)-2.$$

So try again on the closed sub-interval $[-2,e^2-3]\subset(-3,\infty)$.

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Set $f(x)=x^3-\ln(x+3)+2$ and take limits as $x\rightarrow (-3)^+$ and as $x\rightarrow \infty$. If you find that these limits have different signs, then there should be an $a>(-3)$ close to $(-3)$ such that $\mbox{sgn}(f(a))=\mbox{sgn}(\lim_{x\rightarrow (-3)^+}f(x))$, and a $b>a$ such that $\mbox{sgn}(f(b))=-\mbox{sgn}(f(a))$. Then you can use the mean value theorem on the interval $[a,b]$.

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