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Answer the following.
Show that the function f(x) = {x², x ≤ 1; 1/x, x > 1} is continuous at x = 1 but not differentiable.
Show that the function $f(x) = \begin{cases} x^2, & x \le 1 \\ \frac{1}{x}, & x > 1 \end{cases}$ is continuous at $x = 1$ but not differentiable.
Answer
Answer
$f$ is continuous at $x=1$ but $f'(1^-) \ne f'(1^+)$, so it is not differentiable at $x=1$.
- Left limit: $\lim_{x\to 1^-} x^2 = 1$
- Right limit: $\lim_{x\to 1^+} \frac{1}{x} = 1$
- $f(1) = 1^2 = 1$
- Left limit = right limit = $f(1)$, so $f$ is continuous at $x=1$
- Left derivative: $f'(1^-) = \frac{d}{dx}(x^2)\big|_{x=1} = 2$
- Right derivative: $f'(1^+) = \frac{d}{dx}\left(\frac{1}{x}\right)\Big|_{x=1} = -1$
- $2 \ne -1$, so $f$ is not differentiable at $x=1$
- Hence $f$ is continuous but not differentiable at $x=1$