Sketch the graph of f(x) = cases x, |x|< 2,, 1, |x|>2

Sketch the graph of $f(x) = \begin{cases} x, & |x|\lt 2,\\ 1, & |x|>2, \end{cases}$ and explain why Lagrange's theorem cannot be applied to it on $[0,2]$.

Step 1: Where the definition runs out

The two branches cover $|x|\lt 2$ and $|x|>2$. The value $x = 2$ satisfies neither strict inequality, so $f(2)$ is simply not defined — and the same is true at $x = -2$.

Step 2: What Lagrange's theorem requires

The theorem asks that $f$ be continuous on the closed interval $[a,b]$ and differentiable on $(a,b)$; it then produces $\xi\in(a,b)$ with

\[ f'(\xi) = \frac{f(b)-f(a)}{b-a}. \]

Both the hypothesis and the conclusion refer to $f(b)$.

Step 3: Why it cannot be applied here

With $a = 0$ and $b = 2$, the quantity $f(2)$ does not exist. The function is therefore not defined on all of $[0,2]$, let alone continuous there, and the theorem's very first hypothesis fails. The conclusion cannot even be stated, since $\dfrac{f(2)-f(0)}{2}$ is meaningless.

Step 4: Even a repair would not help

Suppose a value were assigned at $x = 2$. The two branches approach different limits there, $\lim_{x\to2^{-}}f(x) = 2, \qquad \lim_{x\to2^{+}}f(x) = 1$.

so $f$ would have a jump discontinuity at $2$ whatever value were chosen, and continuity on $[0,2]$ would still fail unless the value $2$ were assigned — which would make the branch $f(x) = x$ hold on the closed interval and the theorem apply after all, with $f'(\xi) = 1$ for every $\xi$.

Step 5: Verification

\[\text{On } [0,2):\ f(x) = x,\ f'(x) = 1 \text{ throughout}\]

\[\text{On } (2,\infty):\ f(x) = 1,\ f'(x) = 0\]

\[\text{The open circles at } x = \pm2 \text{ mark the four missing values} \;\;\checkmark\]