Lagrange’s Mean Value Theorem

Lagrange’s Mean Value Theorem

Theorem 7.3

Let f (x) be continuous in a closed interval [a, b] and  differentiable in the open interval (a , b) (where f (a), f (b) are not necessarily equal). Then there exist at least one point c ∈( a , b) such that, 

 f ′(c) = f (b) − f (a) / b − a     ... (6)

Remark

If f (a) = f (b) then Lagrange’s Mean Value Theorem gives the Rolle’s theorem. It is also known as rotated Rolle’s Theorem.

Remark

A physical meaning of the above theorem is the number f (b) − f (a) / b − a =  can be  thought of as the average rate of change in f ( x) over (a, b) and f (c) as an instantaneous change.

A geometrical meaning of the Lagrange’s mean value theorem is that the instantaneous rate of change at some interior point is equal to the average rate of change over the entire interval. This is illustrated as follows :

If a car accelerating from zero takes just 8 seconds to travel 200 m, its average velocity for the 8 second interval is 200/8 = 25 m/s. The Mean Value Theorem says that at some point during the travel the speedometer must read exactly 90 km/h which is equal to 25 m/s.

Theorem 7.4

If f ( x) is continuous in closed interval [a , b] and differentiable in open interval (a , b) and if f ′( x) > 0, ∀x ∈ (a, b) , then for, x₁ , x₂ ∈[a , b] , such that x₁ < x₂ we have, f (x₁ ) < f (x₂ ) .

Proof

By the mean value theorem, there exists a c ∈ ( x₁ , x₂ ) ⊂ (a, b) such that,

 f (x₂ ) − f (x₁)  /  x₂ − x₁ = f ′(c)

Since f ′(c) > 0 , and x₂ − x₁ > 0 we have f (x₂) – f (x₁) > 0.

We conclude that, whenever x₁ < x₂ , we have f (x₁ ) < f (x₂ ) .

Remark

If f ′( x) < 0, ∀x ∈ (a, b) , then for, x₁ , x₂ ∈ [a, b] , such that x₁ < x₂ we have, f (x₁ ) > f (x₂ ) .

The proof is similar.

Example 7.25

Find the values in the interval (1,2) of the mean value theorem satisfied by the function f (x) = − x - x² for 1 ≤ x ≤ 2.

Solution

f (1) = 0 and f (2) = −2 . Clearly f ( x) is defined and differentiable in 1 < x < 2 . Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that

f ′(c) = f (2) − f (1) / 2 −1 = 1− 2c

That is, 1− 2c = −2 ⇒ c = 3/2 .

Geometrical meaning

Geometrically, the mean value theorem says the secant to the curve y = f ( x) between x = a and x = b is parallel to a tangent line of the curve, at some point c ∈(a, b) .

Consequences of Lagrange’s Mean Value Theorem

There are three important consequences of MVT for derivatives.

(1) To determine the monotonicity of the given function (Theorem 7.4)

(2) If f ′(x) = 0 for all x in (a, b) , then f is constant on (a, b) .

(3) If f ′( x) = g′( x) for all x , then f ( x) = g( x) + C for some constant C .