Rolle’s Theorem and Lagrange’s Mean Value Theorem

MATMean Value Theorems

These two existence theorems link the behaviour of a function on an interval to its derivative at an interior point. Rolle's theorem is the special case where the endpoint values are equal, guaranteeing a horizontal tangent somewhere inside; Lagrange's Mean Value Theorem (LMVT) generalises this to say the average rate of change equals the instantaneous rate at some interior point.

ISC questions ask you to verify the three hypotheses, then actually solve $f'(c)=$ (the required value) to find $c$ in $(a,b)$ .

Rolle's theorem

f(a)=f(b)\ \Rightarrow\ \exists\, c \in (a,b):\ f'(c)=0

f

is continuous on

[a,b]

, differentiable on

(a,b)

, and

f(a)=f(b)

Lagrange's Mean Value Theorem

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

f

continuous on

[a,b]

and differentiable on

(a,b)

; then such a

c\in(a,b)

exists.

Geometric meaning of LMVT

\text{slope of tangent at } c = \text{slope of chord joining } (a,f(a)),(b,f(b))

c

is the interior point whose tangent is parallel to the chord across

[a,b]

Verification order: first check continuity on $[a,b]$ , then differentiability on $(a,b)$ ; only then apply the conclusion.
Rolle requires the extra equal-endpoint condition $f(a)=f(b)$ ; LMVT does not.
Rolle's theorem is exactly LMVT with $f(a)=f(b)$ , which makes the chord slope $0$ .
Polynomials, $e^x$ , $\sin x$ , $\cos x$ are continuous and differentiable everywhere, so the hypotheses hold on any interval.
After computing $c$ , you MUST confirm it lies in the OPEN interval $(a,b)$ ; a root outside is rejected.
Products like $e^{x}\sin x$ are differentiated with the product rule; for the $[0,\pi]$ verification $f'(c)=0$ gives $\tan c=-1$ , so $c=\dfrac{3\pi}{4}$ .
These theorems are existence results — they guarantee at least one $c$ but do not require it to be unique.

Where the marks go

Skipping the hypothesis check and jumping straight to solving for $c$ — verification of continuity and differentiability is graded.
Accepting a value of $c$ that lies on or outside the interval; $c$ must be strictly interior, i.e. $a<c<b$ .
Confusing the two theorems — using $f'(c)=0$ for an LMVT problem where $f(a)\ne f(b)$ .
Sign and algebra slips when solving $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ , especially with square-root or log functions.

How the board asks it

Derive / proveverify continuity, differentiability, $f(a)=f(b)$ , then solve $f'(c)=0$
Verify Rolle's theorem for $f(x)=x^2-5x+6$ on the interval $[2,3]$ , and find the value of $c$ in the open interval $(2,3)$ for which $f'(c)=0$ .
Numerical $f'(c)=\dfrac{f(b)-f(a)}{b-a}$
Using Lagrange's Mean Value Theorem, find a point $c$ on the curve $f(x)=\sqrt{x^2-4}$ on $[2,4]$ where the tangent is parallel to the chord joining the end points.
Give reasonshypotheses fail (continuity, differentiability or equal endpoints)
Examine whether Rolle's theorem is applicable to $f(x)=|x-1|$ on $[0,2]$ . Give reasons for your answer.
Derive / proveproduct rule on $e^{x}\sin x$ giving $\tan c=-1$
Verify the conditions of Rolle's theorem for $f(x)=e^{x}\sin x$ on $[0,\pi]$ and hence find the value of $c$ at which $f'(c)=0$ .
Derive / proveapply lmvt to $f(x)=\log x$ to bound the mean slope
Using Lagrange's Mean Value Theorem, prove that for $0<a<b$ , $\dfrac{b-a}{b}<\log\dfrac{b}{a}<\dfrac{b-a}{a}$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.