2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Could you please write it rigorously? $\endgroup$ – Ziya Mar 21 '18 at 14:18
  • 3
    $\begingroup$ He wrote the definition, then re-wrote it in symbols. Why do you think it is not rigorous? $\endgroup$ – GEdgar Mar 21 '18 at 14:19
  • 2
    $\begingroup$ I just wanted to know is there any way to write a formal solution? $\endgroup$ – Ziya Mar 21 '18 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.