Show that if $f(-1)<0$ and $f(1)>0$, then there exist $-1<a<1$ such that $f(a)=0$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ continuous. Show that if $f(-1)<0$ and $f(1)>0$, then there exist $-1<a<1$ such that $f(a)=0$.

My approach: Suppose that $f(x)\neq 0$ for all $x\in (-1,1)$. Then $$f((-1,1))=(f(-1),0)\cup(0,f(1))\qquad\qquad(*)$$

Furthermore, $f$ is continuous and the image of connected set is connected, then (*) is a contradiction.

Therefore, must be exist some $a\in(-1,1)$ such that $f(a)=0$. This is correct? Can give me some hint. Thanks!

Answer 1

Just because $f(x) \neq 0$ on $x \in (-1,1)$ does not mean that $0$ is the only omitted value, which you assert via "$(f(-1),0) \cup (0,f(1))$"${} = (f(-1),f(1)) \smallsetminus \{0\}$. My age (in years) is continuous and has never been $-1$, but this does not mean that $-1$ is the only number my age can never be.

Answer 2

Hint: the only connected subsets of $\Bbb{R}$ are intervals (closed, open, half-closed/half-open and possibly with infinite endpoints). The image $f[[-1, 1]]$ of the closed interval $[-1, 1]$ under your function $f$ is connected and includes $f(-1) < 0$ and $f(1) > 0$, so it is an interval containing a negative number and a positive number. Can you take it from there?

Answer 3

(*) is not correct. You are assuming that all values of $f$ are between $f(-1)$ and $f(1)$ which may not be true. Use: $f(-1,1)=[f(-1,1)\cap (-\infty,0)]\cup [f(-1,1)\cap (0,\infty)]$.