Answer
$f$ is both continuous and differentiable at $x=0$.
Work Step by Step
a) The function is graphed below.
c) To test the differentiability of the function at $x=0$, we examine whether the left-hand derivative is the equal to the right-hand one.
- For $x\to0^-$, since $x\lt0$, we have $f(x)=x$. That means $f'(x)=1$.
Therefore, $\lim_{x\to0^-}f(x)=1$
- For $x\to0^+$, since $x\gt0$, we have $f(x)=\tan x$. That means $f'(x)=\sec^2x$.
Therefore, $\lim_{x\to0^+}f(x)=\sec^20=1^2=1$
Since $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)$, $f'(0)=\lim_{x\to0}f(x)$ exists and $f(x)$ is differentiable at $x=0$ as a result.
b) Since $f$ is differentiable at $x=0$, $f$ is also continuous at $x=0$.
