Answer
$f$ is continuous but not differentiable at $x=1$.
Work Step by Step
a) The function is graphed below.
c) To test the differentiability of the function at $x=1$, we examine whether the left-hand limit of the derivative is the equal to the right-hand one at $x=1$.
- For $x\to1^-$, since $x\lt1$, we have $f(x)=x$. That means $f'(x)=1$.
Therefore, $\lim_{x\to1^-}f'(x)=1$
- For $x\to1^+$, since $x\gt1$, we have $f(x)=2-x$. That means $f'(x)=-1$.
Therefore, $\lim_{x\to1^+}f'(x)=-1$
Since $\lim_{x\to1^+}f'(x)\ne\lim_{x\to1^-}f'(x)$, $f'(1)=\lim_{x\to1}f'(x)$ does not exist and $f(x)$ is not differentiable at $x=1$ as a result.
b) To test the continuity of the function at $x=1$, we examine whether the left-hand limit is equal to the right-hand one at $x=1$.
- For $x\to1^-$, since $x\lt1$, we have $f(x)=x$.
Therefore, $\lim_{x\to1^-}f(x)=1$
- For $x\to1^+$, since $x\gt1$, we have $f(x)=2-x$.
Therefore, $\lim_{x\to1^+}f(x)=2-1=1$
Since $\lim_{x\to1^+}f(x)=\lim_{x\to1^-}f(x)$, $\lim_{x\to1}f(x)$ exists and $f(x)$ is continuous at $x=1$ as a result.
