# Give an example of a monotonic increasing function which does not satisfy intermediate value property.

Give an example of a monotonic increasing function which does not satisfy intermediate value property.

I have no idea whether that function would be continuous or discontinuous. Please give me the example(s) with proper reason. Thank you very much.

• Just put in a jump at one point...take $f(x)=x-1$ for $x≤0$ and $f(x)=x+1$ for $x>0$. Or did you want something else? – lulu Apr 9 '17 at 11:40
• @lulu it should not satisfy IVP – rama_ran Apr 9 '17 at 12:03
• @rama_ran That function does not satisfy the IVP. – preferred_anon Apr 9 '17 at 12:10
• My function doesn't satisfy the IVP. For example we have $f(0)=-1$ and $f(1)=2$ so the IVP would tell us that there was a solution to $f(x)=0$ between $0$ and $1$, but there clearly isn't. More simply, just look at the graph. – lulu Apr 9 '17 at 12:27

Well, the intermediate value theorem says that any continuous $f:I \to \mathbb{R}$, where $I \subset \mathbb{R}$ is an interval (and we allow notations like $(-\infty,\infty)=\mathbb{R}$) has the intermediate value property. So in order to find a function which fails to have the intermediate value property, it had better either be
• defined on a set which is not an interval in $\mathbb{R}$
In fact, we'll use essentially the same function for both properties. Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $$f(x)=\cases{0 \quad \text{ if }x<0\\1\quad \text{ if }x>0}$$ and $f(0)=0$. Then $f$ is continuous everywhere except $0$, but it does not have the intermediate value property: $f$ takes the value $0$ (at $-1$), and it takes the value $1$ (at $1$), but there is no point $x$ such that $-1<x< 1$ and $f(x)=\frac{1}{2}$, say.
If, instead of taking $f$ as a function $\mathbb{R} \to \mathbb{R}$, we choose to look at it as a function $\mathbb{R} \setminus \{0\} \to \mathbb{R}$ (and do not define it at $x=0$), then it is continuous at every point at which it is defined, but fails to have the intermediate value property for the same reason. So both hypotheses of the theorem are necessary!