1. Given functions

We need to find the composite function

2. First simplify $f(x)$

For $x\ge 0$,

So

3. Now compute $g(f(x))$ case-wise

Recall:

• if the input to $g$ is negative, then $g(t)=t+1$;
• if the input to $g$ is nonnegative, then $g(t)=1$.

So we must check the sign of $f(x)$.

Case 1: $x<0$

Then

Now:

• if $x+1<0 \iff x<-1$, then $g(f(x))=g(x+1)=(x+1)+1=x+2;$
• if $x+1\ge 0 \iff -1\le x<0$, then $g(f(x))=1.$

Thus for $x<0$,

Case 2: $0\le x<1$

Then

Since $1-x>0$ for $0\le x<1$, we have

Case 3: $x\ge 1$

Then

Hence

Combining all cases,

4. Check continuity

The only possible problematic point is where the formula changes, i.e. at $x=-1$.

• Left-hand limit at $x=-1$: $$\lim_{x\to -1^-}(g\circ f)(x)=\lim_{x\to -1^-}(x+2)=1.$$
• Right-hand limit at $x=-1$: $$\lim_{x\to -1^+}(g\circ f)(x)=1.$$
• Function value: $$(g\circ f)(-1)=1.$$

So $(g\circ f)(x)$ is continuous at $x=-1$.

Everywhere else each piece is continuous, hence $(g\circ f)(x)$ is continuous everywhere.

So option C is false.

5. Check differentiability

For $x<-1$,

For $x>-1$,

At $x=-1$:

• left derivative $=1$
• right derivative $=0$

These are unequal, so $(g\circ f)(x)$ is not differentiable at $x=-1$.

At all other points it is differentiable.

Thus $(g\circ f)(x)$ is:

• continuous everywhere,
• not differentiable at exactly one point.

6. Evaluate options

• A: continuous everywhere but not differentiable at $x=1$
  False, the non-differentiable point is $x=-1$.

• B: differentiable everywhere
  False.

• C: not continuous at $x=-1$
  False.

• D: continuous everywhere but not differentiable exactly at one point
  True.

Therefore, the correct answer is