Does $f(x)=\begin{cases}x+1& 0I am trying to solve the problem below but I think my work is not so convincing to me. Here is the problem:
$$f(x)=\begin{cases}x+1& 0<x\leq 1,\\0 &\text{if}\;\; x=0.\end{cases}$$
Show that $f$ does not satisfy intermediate value property.
Here is my work:
Solution
If $x$ is continuous on $[0,1]$ where $\alpha, \beta\in[0,1]$ with $\alpha< \beta$. Also, if $y$ is an intermediate point between $f(\alpha)$ and $f(\beta)$, i.e., $f(\alpha)<y<f(\beta)$ or $f(\beta)<y<f(\alpha)$, then $\exists\;c\in[\alpha,\beta]$ s.t. $f(c)=y.$
Now, let $\alpha, \beta\in[0,1]$ be arbitrary s.t.  $f(\alpha)<f(\beta)$. Then, we find a $c\in[\alpha,\beta]$ s.t. $f(c)=y.$
In particular, let $\alpha=0\;\&\;\; \beta=1$. Clearly, $\alpha, \beta\in[0,1]$. Then, $f(\alpha)=0$ and $f(\beta)=2$. Let $y=0.5$ be an intermediate point between $f(\alpha)$ and $f(\beta)$, then $f(0)<y<f(2)$. But $\nexists c\in[0,1]$ s.t. $f(c)=0.5$.
Hence, $f$ does not satisfy IVP. Can anyone help check if this is correct? Better solutions are highly needed!
 A: The intermediate value theorem is not usable when you set $\alpha =0$ because $f:[0,r)\to\mathbb{R}$ with $r>0$ is not continuous in $x=0$. For any $1>\varepsilon>0$ and any $\delta>0$ you get $|0-(0+\delta)|\leq \delta$ but $$|f(0)-f(0+\delta)|=|0-(\delta+1)|=1+\delta>\varepsilon+\delta>\varepsilon$$
It really does not hold for $\alpha=0$, since for any $y\in(0,1]$ we cannot find a $c\in(0,\beta)$ (with $\beta < r$) with $f(c)=y$, because $f(c)=1+c>y$ (since $c>0$).
On the other hand, it does indeed hold as soon as soon as $\alpha>0$ because $f:(0,r)\to\mathbb{R}$ is just $f(x)=x+1$. So given $0<\alpha<\beta$ and $f(\alpha)<y<f(\beta)$ we can set $c=y-1$ and, since $\alpha=f(\alpha)-1$ and $\beta=f(\beta)-1$, get $\alpha<c<\beta$, or $c\in(\alpha,\beta)$.
If we take $r=1.1$, $\alpha=0.1$ and $\beta=1$, we get indeed $f(\alpha)=1.1$ and $f(\beta)=2$. With $y=1.5$, we have to choose $c=0.5$ (as described above), and have $f(0.1)<y<f(2)$ as well as $c\in(0,1]$.
Your mistake was that for whatever reason you weren't able to find a $c$. But, and that is important, one's inability to find an object satisfying a certain property does not imply that there is none. On the contrary, only if we can proof that no object can possibly satisfy the property, we can speak of non existence.
A: "In particular, let α=0.1&β=1. Clearly, α,β∈(0,1]. Then, f(α)=1.1 and f(β)=2. Let y=1.5 be an intermediate point between f(α) and f(β), then f(0.1)

Um.... What about $c = 0.5$?  $f(0.5) = 1.5$ and $0.5 \in (0,1]$.  
As $f$ is continuous on $(0,1]$, the IVT WILL hold.  So it seems silly you went out of you're way to avoid $x = 0$ when $f$ is not continuous at $x = 0$.  The IVT can only fail when $f$ is NOT continuous.  And as $f$ is continuous on $(0,1]$ but not at $x=0$ to find a counterexample we will have to consider that $f(0)=0$ but $f(a)> 1$ for $a > 0$.  SO there is no $c \in [0,a]$ so that $0 < f(c)< 1$.
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Far too wordy and obtuse.  And you claim that there is no $c$ so that $f(c) = \frac 12$.  How do you know that?  I'd be a little more willing to take you word for it if you hadnt be so needlessly detailed in restating the exact statement of the IVT.
Simply state:
$f(0) = 0$ and $f(1) =2$ and $0 < \frac 12 < 1$.  If the IVT held, there would exist a $c$ so that $0< c < 1$ and $f(c) = \frac{1}{2}$.
But if $0 < c < 1$ then $f(c) = c+1$ and $\frac 12 < 1 < c+1 < 2$ so $c+1$ can not be equal to $\frac 12$.  So the IVT does not hold.
That is all you need.
For extra Brownie points, it'd be nice to point out: The IVT holds on intervals where $f$ is continuous.  $f$ is not continuous at $x = 0$ so if fails, it fails when we consider that $0 = f(0) \ne \lim_{x\to 0^+} f(x) = 1$.
