# Implementation Of The Intermediate Value Theorem

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?

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 $$\varphi(c)=0$$.

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 $$[0,M]$$.

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