The Secant Meets Its Parallel Tangent

Somewhere on the curve sits a tangent parallel to the secant. This scene lets you hunt for that point.

The secant's slope is the average rate of change. At some interior point the instantaneous slope matches it exactly. That is what the mean value theorem guarantees.

For example, a πŸš— car covering 100 km in one hour averages 100 km/h. At some moment in that hour, the speedometer read exactly 100.

fβ€²(ΞΎ)=f(b)βˆ’f(a)bβˆ’af'(\xi) = \frac{f(b)-f(a)}{b-a}

Part of these learning paths

Examples & Exercises

Apply the Mean Value Theorem to f(x)=x2f(x) = x^2 on [0,2][0, 2] and find ΞΎ\xi explicitly.

f(2)βˆ’f(0)2βˆ’0=2,fβ€²(ΞΎ)=2ΞΎ=2β€…β€Šβ‡’β€…β€ŠΞΎ=1\frac{f(2)-f(0)}{2-0} = 2, \quad f'(\xi) = 2\xi = 2 \;\Rightarrow\; \xi = 1
f(x)=x2f(x) = x^2, ΞΎ=1\xi = 1 β€” the tangent slope is 2, matching the secant.
Show proof

  1. The secant slope is f(2)βˆ’f(0)2βˆ’0=4βˆ’02=2\frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2} = 2.

  2. Since fβ€²(x)=2xf'(x) = 2x, fβ€²(ΞΎ)=2f'(\xi) = 2 gives 2ΞΎ=22\xi = 2, i.e. ΞΎ=1\xi = 1. As 1∈(0,2)1 \in (0, 2), this matches the theorem's conclusion.

When f(a)=f(b)f(a) = f(b), the Mean Value Theorem becomes Rolle's theorem.

Check it for f(x)=x3βˆ’3xf(x) = x^3 - 3x on [βˆ’1,2][-1, 2].

f(βˆ’1)=f(2)=2β€…β€Šβ‡’β€…β€Šβˆƒβ€‰ΞΎβˆˆ(βˆ’1,2):fβ€²(ΞΎ)=0f(-1) = f(2) = 2 \;\Rightarrow\; \exists\,\xi \in (-1,2): f'(\xi) = 0
At ΞΎβ‰ˆ1\xi \approx 1 the tangent is horizontal β€” parallel to the secant (slope 0).
Show proof

  1. The function ff is a polynomial, hence continuous and differentiable everywhere; in particular it is continuous on [βˆ’1,2][-1, 2] and differentiable on (βˆ’1,2)(-1, 2), so the hypotheses of Rolle's theorem and the MVT are satisfied. f(βˆ’1)=βˆ’1+3=2f(-1) = -1 + 3 = 2 and f(2)=8βˆ’6=2f(2) = 8 - 6 = 2, so the secant slope is 2βˆ’23=0\frac{2-2}{3} = 0.

  2. fβ€²(x)=3x2βˆ’3=0f'(x) = 3x^2 - 3 = 0 gives x=Β±1x = \pm 1, and the open interval (βˆ’1,2)(-1, 2) contains ΞΎ=1\xi = 1. That is Rolle's theorem β€” the Mean Value Theorem with zero slope.

Prove via the Mean Value Theorem that f′≑0f' \equiv 0 forces ff to be constant.

f(b)βˆ’f(a)=fβ€²(ΞΎ)(bβˆ’a)=0f(b) - f(a) = f'(\xi)(b-a) = 0
A constant function: both secant and tangent have slope 0.
Show proof

  1. The hypothesis is that fβ€²=0f' = 0 at every point: assume ff is continuous on every interval in its domain, differentiable on the interior, with f′≑0f' \equiv 0 everywhere.

  2. Take any two points u<vu < v. The hypothesis is inherited by [u,v][u, v], so by the Mean Value Theorem some ξ∈(u,v)\xi \in (u, v) gives f(v)βˆ’f(u)=fβ€²(ΞΎ)(vβˆ’u)=0β‹…(vβˆ’u)=0f(v) - f(u) = f'(\xi)(v-u) = 0\cdot(v-u) = 0.

  3. Hence f(v)=f(u)f(v) = f(u) for every pair u<vu < v, so ff is constant.

10. Differentiation