Intermediate Value Theorem.

Question

Let $f:[a,b] \to \mathbb{R}$ be a continuous function and let $f(a)<f(b)$. Then by intermediate value theorem

$(1) f([a,b])=[f(a),f(b)]\\$

$(2)f([a,b])\supseteq [f(a),f(b)]\\ $

$(3) f([a,b])\subseteq [f(a),f(b)]\\$

$(4) f([a,b])\ne [f(a),f(b)]$

For example if I take the function $f(x)=x^2$ on $[-1,2]$ then $f([-1,2])=[0,4]$ which is not equal to $[f(-1),f(2)]=[1,4]$, also $[1,4]\subseteq[0,4]$, so options $1$ is incorrect. But how to use IVT here? Please help.

Answer 1

Let $c\in[f(a),f(b)]$ then by IVT (since $f$ is continuous) it follows that there exists some $x\in[a,b]$ such that $f(x)=c$. Therefore $c\in f([a,b])$ (the image of interval $[a,b]$ under $f$). Thus $$[f(a),f(b)]\subseteq f([a,b])$$

Answer 2

IVT says that $f$ take all value between $f(a)$ and $f(b)$. In other word that $$[f(a),f(b)]\subseteq f([a,b]).$$