Season 2 Episode 4

In this episode, Aditi introduces the fundamentals of differentiation and integration, as they're taught on a university Maths degree. We take a look at continuous and differentiable functions, and prove the fundamental theorem of calculus.

Further Reading

Continuous, differentiable or neither?

Here are some more functions. Have a go at sketching them, and decide whether they are continuous, differentiable, or neither.

$$f(x)=x \quad \text{if $x\geq0$}, \quad f(x)=0 \quad \text{if $x<0$}$$

$$f(x)=x^2-x \quad\text{if $x\geq 2$}, \quad f(x)=-x^2+7x-8 \quad \text{if $x < 2$}$$

$$f(x)=\left(x\quad\text{rounded to 3 decimal places}\right)$$

Rolle’s Theorem

We skipped a bit of the discussion of Rolle’s Theorem for time- here’s something to think about.

Theorem: If a function $f(x)$ is continuous and differentiable, and it has $f(a)=f(b)$, then there’s a point $c$ in between where $f’(c)=0$.

Discussion: The proof of this is in the first year of most university mathematics courses, and it’s not conceptually tricky, but it is a bit fiddly. Essentially, we look at the maximum and the minimum of the function and show that one or both of them have $f’(c)=0$ by thinking about the behaviour of $f(x)-f(c)$ near those points. It’s made complicated by the fact that (in general) a set of values doesn’t have to have a maximum, even if all the values are bounded below some number. For example, the set $\lbrace\frac{1}{2} , \frac{2}{3},\frac{3}{4},\frac{4}{5},\dots\rbrace$ has all its values below 1, but there isn’t a maximum “biggest” element! In the proof for Rolle’s Theorem, we use the fact that the function is continuous to prove that it actually achieves a maximum value (rather than just getting close to some upper limit).

If you’re interested in the technical details, there’s a proof of Rolle’s theorem on Wikipedia which uses the Extreme Value Theorem, which uses the Bolzano-Weierstrass theorem, which uses the Monotone Convergence Theorem. A university course in Analysis would usually cover all of these in the opposite order, so that you build from basic theorems to more advanced theorems.

Exercise: Explore why the conditions in Rolle’s Theorem are there:

• Find a function $f(x)$ that is continuous and differentiable with $f(a)\neq f(b)$ and which doesn’t have any point $c$ in between where $f’(c )=0$.
• Find a function $f(x)$ that is continuous but not differentiable with $f(a)=f(b)$ and which doesn’t have any point $c$ in between where $f’(c )=0$.

But the conditions are not, strictly speaking, necessary:

• Find a function $f(x)$ that is continuous and differentiable with $f(a)\neq f(b)$, but which happens to have a point $c$ in between where $f’(c)=0$.
• Find a function $f(x)$ that is continuous but not differentiable with $f(a)=f(b)$, but which happens to have a point $c$ in between where $f’(c)=0$. 

Mean Value Theorem slo-mo replay

On the stream, we used Rolle’s Theorem above to prove the Mean Value Theorem. Here’s a step-by-step replay!

Theorem: if a function $f(x)$ is continuous and differentiable, then there’s a point $c$ in between where $\displaystyle f’(c )=\frac{f(b)-f(a)}{b-a}$.

Proof: We can’t use Rolle’s Theorem straight away on $f(x)$ because we don’t have the $f(a)=f(b)$ condition. Instead, we’ll consider the function

$$ g(x)= f(x)-\left[f(a)+\left(\frac{f(b-f(a)}{b-a}\right)(x-a)\right]$$

(Looks like magic, but Aditi explained on the livestream how this comes from the idea of rotating the graph around.)

Then $g(x)$ is continuous and differentiable and has $g(a)=g(b)$ (check it – it’s zero at both $a$ and $b$).

Those are the conditions we need for Rolle’s theorem for $g(x)$. So by Rolle’s Theorem there’s a point in between with $g’(c )=0$. What does $g’(c )=0$ actually mean? Well, if we differentiate $g(x)$ we get

$$g’(x)=f’(x)-\left(\frac{f(b-f(a)}{b-a}\right).$$

So a point with $g’(c )=0$ would also have $\displaystyle f’(c )=\frac{f(b)-f(a)}{b-a}$. That’s exactly what we want for $f(x)$ so we’re done.

Curve sketching

We’ve seen $\sin\left(\frac{1}{x}\right)$ and $x\sin\left(\frac{1}{x}\right)$ and $x^2\sin\left(\frac{1}{x}\right)$. I’d like you to sketch

$$y=x^{1/2} \sin \left(\frac{1}{x} \right),\quad \text{and}\quad y=x^{3/2} \sin\left(\frac{1}{x}\right)$$

on separate axes. Check your plots against Desmos. Do you think these functions are continuous? Are they differentiable?

If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk.