# True or False: Continuous Functions (Extreme Value Theorem)

Do my justifications seem appropriate?

a.) Every function $f:[0,1]\rightarrow \mathbb{R}$ has a maximum.

True; if $f:[0,1]\rightarrow \mathbb{R}$, $f$ wil be closed and bounded above, so it will have a maximum.

b.) Every continuous function $f:[a,b]\rightarrow \mathbb{R}$ has a maximum.

True; if $f$ is continuous on a closed, bounded interval, then it will have a min and a max.

c.) Every continuous function $f:(0,1)\rightarrow \mathbb{R}$ has a maximum.

False; $f$ is not on a closed interval.

d.) Every continuous function $f:(0,1)\rightarrow \mathbb{R}$ has a bounded image.

False; if $f$ has a vertical asymptote at the $0$ endpoint, the image will be unbounded.

e.) If the image of a continuous function $f:(0,1)\rightarrow \mathbb{R}$ is bounded below, then the function has a minimum.

False; only if the inf$f(D)$ is a functional value.

• Counterexample for a): $x\mapsto \dfrac 1 x$, if $x\neq 0$ and give any value at $0$. Works for c) and d) too. A counterexample for e) is $x\mapsto -x^2$. Commented Oct 30, 2013 at 19:14
• Ah, I see. I was thinking $f$ was continuous in a), but it doesn't have to be. Thank you. Commented Oct 30, 2013 at 19:18

Your part (a) is wrong because $f$ is not assumed to be continuous. You can cook up as nasty a function as you want as a counterexample.
For parts (c), (d), and (e), you should probably write down counterexamples. For instance, in (c), you note: "$f$ is not on a closed interval." This is begging the question, because that's more or less what you're being asked to justify. There are many continuous functions on an open interval that are bounded, so you should explicitly produce a function on an open interval that is not bounded.