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  1. Let $f : \mathbb R \rightarrow \mathbb R $ be a continuous function, such that $$f(1) = 3$$ $$\lim_{x \to \infty} f(x) = -3$$ Prove that there exists at least one $c\in \mathbb R$ such that $f(c) = c$.
  2. Let $f(x)$ be a continuous function on $\mathbb R$, that is bounded for $x ≥ 0$ and non-positive for $x < 0$. Prove that the equation $$f(x)+7x=17$$ has at least one solution.

I know that I need to use the IVT, but I am also looking for the intuition behind solving questions of this nature. I am able to solve general problems with IVT if I know the function.

I have found a solution for a. For b I am claiming that $$ x < 0 \rightarrow g(x) < 0 $$ $$ \lim_{x \to \infty} g(x) = \infty> 0$$can I use the IVT on an interval $[0,\infty)$ , or it must be a closed interval?

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Let $\varphi(x)=f(x)-x$ and we have $\lim_{x\rightarrow\infty}\varphi(x)=-\infty$ and $\varphi(1)=3-1=2>0$, so some $c$ is such that $\varphi(c)=0$.

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In the first case, you can (almost) apply apply the theorem to the function $g(x)=f(x)-x$. You know that $g(1)=2$ and that $\lim_{x\to\infty}g(x)=-\infty$. So, applying the theorem to $g$, there is some $c\in(1,\infty)$ such that $g(c)=0$.

For the other problem, take $g(x)=f(x)+7x$ and do something similar.


Concerning your solution of the second problem, no, you should not apply the intermediate value theorem directly to infinite intervals. However, since $\lim_{x\to\infty}g(x)=\infty$, there is a $M\in\mathbb R$ such that $g(M)>17$. Now, apply the intermediate value theorem to the interval $[0,M]$.

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  • $\begingroup$ I am sure that you mean $g (1)=2. $ $\endgroup$ – Fred Dec 16 '19 at 19:02
  • $\begingroup$ Sure you're sure! I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Dec 16 '19 at 19:16
  • $\begingroup$ How should I use the bounded statement to my advantage ? $\endgroup$ – johnadams12 Dec 16 '19 at 19:40
  • $\begingroup$ Since $f|_{[0,\infty)}$ is bounded and $\lim_{x\to\infty}7x=\infty$, $\lim_{x\to\infty}g(x)=\infty$. $\endgroup$ – José Carlos Santos Dec 16 '19 at 19:44
  • $\begingroup$ I have edited with a proposed solution , can you make sure I am on the right track ? $\endgroup$ – johnadams12 Dec 16 '19 at 20:00

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