Given function

$f(x)=\min\{x+1, |x|+1\}$

We first simplify it by considering cases.

Case 1: $x\ge 0$

For $x\ge 0$, we have $|x|=x$ so $x+1=|x|+1=x+1$ Hence, $f(x)=\min\{x+1,x+1\}=x+1$

Case 2: $x<0$

For $x<0$, we have $|x|=-x$ so the two expressions are $x+1 \quad \text{and} \quad -x+1$ Now compare them: $x+1< -x+1 \iff 2x<0$ which is true for $x<0$.

Therefore, $f(x)=x+1$

Thus for all real $x$

$f(x)=x+1$

So the given function is simply a linear function.

Differentiability

Since $f(x)=x+1$ for all $x\in\mathbb R$, it is differentiable everywhere, with $f'(x)=1$

Check options

• A: $f(x)$ is differentiable everywhere. ✅ True
• B: $f(x)$ is not differentiable at $x=0$. ❌ False
• C: $f(x)>1$ for all $x\in\mathbb R$. ❌ False, because for example at $x=-1$, $f(-1)=0$
• D: $f(x)$ is not differentiable at $x=1$. ❌ False

Conclusion

The correct option is: $\boxed{A}$

Comparison with stored answer

Stored correct answer: A

My derived answer also is A, so they agree.