# Lagrange’s Mean Value Theorem

Here you will learn lagrange’s mean value theorem statement, its geometrical and physical interpretation with examples.

Let’s begin –

## Lagrange’s Mean Value Theorem (LMVT)

Statement : Let f be a function that satisfies the following conditions :

(i) f is continuous in [a, b]

(ii) f is differentiable in (a, b)

Then there is a number c in (a, b) such that f'(c) = $$f(b) – f(a)\over b – a$$

#### Geometrical Interpretation :

Geometrically, the lagrange’s mean value theorem says that somewhere between A and B the curve has atleast on tangent parallel to chord AB.

#### Physical Interpretation :

If we think  of the number (f(b) – f(a))/(b – a) as the average change in f over [a, b] and f'(c) as an instantaneous change, then the mean value theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.

Example : find c of the Lagranges mean value theorem for the function f(x) = $$3x^2$$ + 5x + 7 in the interval [1, 3].

Solution : Given f(x) = $$3x^2$$ + 5x + 7

$$\implies$$ f(1) = 3 + 5 + 7 = 15 and f(3) = 27 + 15 + 7 = 49

Now, Differentiating f(x) with respect to x,

$$\implies$$ f'(x) = 6x + 5

Here a = 1, b = 3

Now from lagrange’s mean value theorem

f'(c) = $$f(b) – f(a)\over b – a$$ $$\implies$$ 6c + 5 = $$49 – 15\over 2$$ = 17

$$\implies$$ c = 2

Example : If f(x) is continuous and differentiable over [-2, 5] and -4 < f'(x) < 3 for all x in (-2, 5), then the greatest possible value of f(5) – f(-2) is –

Solution : Applying Lagranges mean value theorem (LMVT),

f'(x) = $$f(5) – f(-2)\over 5 -(-2)$$ for some x in (-2, 5)

Now, -4 $$\le$$ $$f(5) – f(-2)\over 7$$ $$\le$$ 3

-28 $$\le$$ f(5) – f(-2) $$\le$$ 21

$$\therefore$$  Greatest possible value of f(5) – f(-2) is 21.