Given f(x) = cases 1+x, x < 0,, x, 0 < x < 1,, 2-x, 1 < x

Given $f(x) = \begin{cases} 1+x, & x \lt 0, \\ x, & 0 \lt x \lt 1, \\ 2-x, & 1 \lt x \lt 2, \\ 3x-x^{2}, & x \gt 2, \end{cases}$ 1° sketch the graph; 2° for which $x$ is $f$ continuous? 3° for which $x$ does $f'(x)$ exist? 4° for which $x$ is $f'$ continuous?

Step 1: The graph

Note first that the definition uses strict inequalities, so $f$ is not assigned any value at $x = 0$, $1$ or $2$. Open circles mark the missing endpoints.

Reading left to right: a line of slope $1$ rising to height $1$; a jump down to $0$; a line of slope $1$ up to height $1$; a line of slope $-1$ down to $0$; a jump up to $2$; then the falling arc of the parabola $3x-x^{2}$.

Step 2: Continuity (2°)

Inside each open piece $f$ is a polynomial, hence continuous. Only the three junctions need checking.

\[x = 0:\quad \lim_{x\to 0^{-}}(1+x) = 1, \qquad \lim_{x\to 0^{+}}x = 0 \;\Longrightarrow\; \text{jump}\]

\[x = 1:\quad \lim_{x\to 1^{-}}x = 1, \qquad \lim_{x\to 1^{+}}(2-x) = 1 \;\Longrightarrow\; \text{limits agree}\]

\[x = 2:\quad \lim_{x\to 2^{-}}(2-x) = 0, \qquad \lim_{x\to 2^{+}}\bigl(3x-x^{2}\bigr) = 2 \;\Longrightarrow\; \text{jump}\]

So $f$ is continuous at every point of its domain $\mathbb{R}\setminus\{0,1,2\}$. At $x = 1$ the discontinuity is removable: setting $f(1) = 1$ makes $f$ continuous there. At $x = 0$ and $x = 2$ the jumps are genuine and no assignment of a value can repair them.

Step 3: Existence of $f'$ (3°)

On each open piece $f$ is differentiable, with

\[ f'(x) = \begin{cases} 1, & x \lt 0, \\ 1, & 0 \lt x \lt 1, \\ -1, & 1 \lt x \lt 2, \\ 3-2x, & x \gt 2. \end{cases} \]

At $x = 0$ and $x = 2$ the function is discontinuous, and a function that is not continuous at a point cannot be differentiable there. At $x = 1$, taking the repaired value $f(1) = 1$, the one-sided derivatives are

\[f'_{-}(1) = \lim_{h\to 0^{-}}\frac{(1+h)-1}{h} = 1, \qquad f'_{+}(1) = \lim_{h\to 0^{+}}\frac{\bigl(2-(1+h)\bigr)-1}{h} = \lim_{h\to 0^{+}}\frac{-h}{h} = -1.\]

They are finite but unequal, so $f'(1)$ does not exist: the graph has a corner at the peak. Hence $f'$ exists exactly on $\mathbb{R}\setminus\{0,1,2\}$.

Step 4: Continuity of $f'$ (4°)

On its domain $f'$ is piecewise constant or linear, hence continuous. It is worth looking at what happens at the three excluded points, because the three cases differ: \[x = 0:\quad \lim_{x\to 0^{-}}f' = 1 = \lim_{x\to 0^{+}}f' \;\Longrightarrow\; \text{the limits agree}\]

\[x = 1:\quad \lim_{x\to 1^{-}}f' = 1, \qquad \lim_{x\to 1^{+}}f' = -1 \;\Longrightarrow\; \text{genuine jump in } f'\]

\[x = 2:\quad \lim_{x\to 2^{-}}f' = -1, \qquad \lim_{x\to 2^{+}}\bigl(3-2x\bigr) = -1 \;\Longrightarrow\; \text{the limits agree}\]

So $f'$ is continuous on $\mathbb{R}\setminus\{0,1,2\}$. At $x = 0$ and $x = 2$ the derivative is undefined only because $f$ itself is; its one-sided limits match, so the discontinuity of $f'$ there is removable. Only at $x = 1$ does $f'$ genuinely jump.