1. **Problem statement:** Construct a numerical function that is unbounded, discontinuous everywhere, but differentiable at exactly one point. 2. **Key concepts:** - A function is **unbounded** if it can take arbitrarily large values. - A function is **discontinuous everywhere** if it has no points where it is continuous except possibly isolated points. - Differentiability at a point requires continuity at that point. 3. **Approach:** We want a function that is discontinuous everywhere except one point where it is differentiable. 4. **Example function:** Consider the function $$f(x) = \begin{cases} x^2 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$$ 5. **Check properties:** - **Unbounded:** As $x \to \infty$ through rationals, $f(x) = x^2$ becomes arbitrarily large, so unbounded. - **Discontinuous everywhere except possibly at 0:** For any $x \neq 0$, the limit from rationals is $x^2$ but from irrationals is 0, so no continuity. - At $x=0$, $f(0) = 0$ for both rationals and irrationals, so continuous at 0. 6. **Differentiability at 0:** Calculate the derivative at 0: $$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$$ - For rational $h$, $f(h) = h^2$, so $\frac{f(h)}{h} = h \to 0$. - For irrational $h$, $f(h) = 0$, so $\frac{f(h)}{h} = 0$. Since both limits agree and equal 0, $f$ is differentiable at 0 with $f'(0) = 0$. 7. **Conclusion:** The function is unbounded, discontinuous everywhere except at 0, and differentiable only at 0. **Final answer:** $$f(x) = \begin{cases} x^2 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$$