Removing the secant, a path from Rolle to the Mean Value Theorem

2026-04-05 602 words 3 minutes

Contents

We look at what it means to subtract the linear trend from a function, why the secant line is the natural choice, and how this simple idea unlocks the Mean Value Theorem from Rolle’s theorem.

You’re probably well acquainted with the $\sin(2x)$ function.

What transformation will transform our traditional $\sin(2x)$ into this:

$\sin(2x) + 0.4x$ in blue, its linear trend $0.4x$ in red, and $\sin(2x)$ recovered in teal after removing the trend.

We add a linear trend of $0.4x$ to $\sin(2x)$ which suddenly makes $\sin(2x)$ not only changing periodically but also increasing. And when we remove that linear trend we get back the original $\sin(2x)$, centered on the $x$-axis.

Consider adding this other linear trend $0.8x$:

A steeper trend $0.8x$ tilts the oscillation more dramatically, but subtracting it again recovers $\sin(2x)$.

Note that in both cases the linear trend acts as the secant line of the function and when we remove this linear trend the function goes back to being centered on the $x$-axis.

Can we generalise this to other functions?

Here’s a quadratic function $f(x) = x^2$ on the interval $[0, 3]$. The secant will be the only straight line that passes through those two endpoints $(0, 0)$ and $(3, 9)$.

$f(x) = x^2$ with its secant through $(0, 0)$ and $(3, 9)$.

And removing it from the function will remove the linear trend of that function. Think of it as decomposing your function into a linear part and the remainder:

$$f(x) = L(x) + g(x)$$

where $L$ is the secant line and $g = f - L$.

The original $f$ (blue), the secant $L$ (red dashed), and $g = f - L$ (teal) on the same axes. Notice $g$ is anchored to zero at both endpoints.

The $x-a$ is a horizontal shift necessary to anchor the line at $a$. When $a=0$,

$$L(x) = f(a) + \frac{f(b) - f(a)}{b - a} \cdot (x-a) = f(a) + \frac{f(b) - f(a)}{b - a} \cdot x$$

but if the interval was $[1,3]$ instead, you can now see the vertical shift accomplished by the factor $x-a$ proves useful:

On $[1,3]$: the correct secant $L$ (using $x-a$) passes through both endpoints, while the false version (using $x$) does not.

This proves useful when solving the Mean Value Theorem from Rolle’s theorem

Rolle’s theorem states: let $f$ be continuous on $[a, b]$, differentiable on $(a, b)$, and suppose $f(a) = f(b)$. Then there exists $c \in (a, b)$ such that

$$f'(c) = 0.$$

While the MVT says: let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists $c \in (a, b)$ such that

$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$

Do you recognise the problem? We want to use Rolle’s theorem to prove the MVT but Rolle’s assumes $f(a) = f(b)$ while the MVT doesn’t. If we remove the secant line or that linear component, suddenly the new transformed function is anchored on the $x$-axis and allows us to apply Rolle’s theorem.

The proof

Define the secant line through $(a, f(a))$ and $(b, f(b))$:

$$L(x) = f(a) + \frac{f(b) - f(a)}{b - a}(x - a)$$

and set $g(x) = f(x) - L(x)$. Since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, so is $g$. We check the endpoints:

$$g(a) = f(a) - L(a) = 0$$$$g(b) = f(b) - L(b) = 0$$

So $g(a) = g(b) = 0$. All the conditions of Rolle’s theorem are satisfied, so there exists $c \in (a, b)$ with $g'(c) = 0$. Differentiating $g$:

$$g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}$$

Setting $g'(c) = 0$ gives:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

which is exactly what the MVT claims. $\square$