Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 14/Solution 1

For any nonzero $x$ that is not of the form $\frac{1}{n}$, the function is differentiable. Indeed, one can find a small interval around such an $x$ that does not contain any points of the form $\frac{1}{n}$, so the function is identically zero in that interval. Hence, $$f^{\prime }(x)=0.$$

For any $a=\frac{1}{n}$, the function is not continuous because $$\underset{x\to a}{\lim }f(x)=0\ne f(a)=\frac{1}{n^2}.$$ Therefore, $f$ is not differentiable at such points.

It remains to consider $a=0$. Using the definition of the derivative, $$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f(h)-f(0)}{h}.$$

If $h=\frac{1}{n}$, then $$\frac{f(h)-f(0)}{h}=\frac{h^2}{h}=h.$$

If $h$ is not of the form $\frac{1}{n}$, then $$\frac{f(h)-f(0)}{h}=0.$$

In either case, the limit as $h\to 0$ is 0. Thus, $$f^{\prime }(0)=0.$$

Therefore, $f$ is differentiable for all $x\ne \frac{1}{n}$, and $$f^{\prime }(x)=0\text{for all such }x.$$