Mean Value Theorem

Mean Value theorem explains whether a function is continuous and differentiable in both the open interval (a, b) and the closed interval [a, b] which directly gives us the signal whether a derivative exists are not. In other words, it gives us information about the velocity of a particle. Suppose  a particle is traveling at 120 km/hr, then its mean speed at some point of time would have been 120 km/hr. This tells us that the path of the curve is continuously differentiable.

Mean Value Theorem For Derivatives

Lagrange’s Mean Value Theorem:

 Let f(x) be a R- valued function such that

(i) f(x) is continuous on [a,b],
(ii) f(x) is differentiable on (a,b).
then there exists a  number c є (a,b) such that f ‘(c) =$\frac{f(b)-f(a)}{b-a}$ .
When geometrically interpreted, the conclusion of the theorem is that point P on the curve, the tangent of which is parallel to the chord AB joining the extremities of the curve.

Let P be the point [c, f(c) ] on the curve such that,
 $\frac{f(b)-f(a)}{b-a}$ = f ‘(c)

Mean Value Theorem for Derivatives

The slope of the chord is AB = $\frac{f(b)-f(a)}{b-a}$  and that of the tangent at P( c, f(c) ] is f ‘ (c). These being equal, it shows that there exists a point P on the curve, such that the tangent lies parallel to the chord AB.

[ f(b) – f(a) ] is the change in the function f as x changes from a to b so that is the average rate of change of the function over the interval [a, b ]. Also f ‘(c) is that actual rate of change of the function for x -> c. Thus, the theorem states that the average rate of change of a function over the interval is also the actual rate of change of the function at some point of the interval.  For instance, the average velocity of a particle over an interval of time is equal to the velocity at some instant belonging to the interval.
This gives  the theorem  the name “Mean Value”.

Mean Value Theorem Examples

Example 1.
f(x) = x2 - 4x  on [2,4]

Solution:

Given f(x) = x2 - 4x  on [2,4]
=>f(2) = 22 - 4 $\times$ 2= 4 - 8 =  -4
=>f(4) = 42 - 4 $\times$ 4 = 0

Using mean value theorem
f'(c) = $\frac{f(b)-f(a)}{b-a}$

=>f'(x) = $\frac{0-(-4)}{4-2}$

=>f'(x)= $\frac{4}{2}$

=>f'(x)=2

Differentiate equation
=>f(x) = x2 - 4x
=>f'(x) = 2x - 4
=>2 = 2x - 4
=>2x = 6
=>x = 3

Example 2.
f(x) = $x^{2}$ - 4x + 7 -1 = x = 3

Solution:


Given f(x) = $x^{2}$ - 4x + 7 -1 = x = 3
=>f(-1) = $(-1)^{2}$ - 4(-1) + 7 = 12
=>f(3) = $3^{2}$ -4  $\times$ 3 + 7 = 4

Using mean value theorem
f'(c) = $\frac{f(b)-f(a)}{b-a}$
=>f'(x) = $\frac{4-12}{3-(-1)}$
=>f'(x) = -2

Differentiate equation
=> f(x) = $x^{2}$ - 4x + 7
=> f'(x) = 2x - 4
=> -2 = 2x - 4   
=> 2x = 2
=> x = 1

Mean Value Theorem for Integrals


If f is continuous on [a, b], then there exists a number c in [a, b] such that,

f(c) = ${f}_{ave}$ = $\frac{1}{b-a}$ $\int_{a}^{b}{f}(x)dx$ 

( i. e )   $\int_{a}^{b}{f}(x)dx$= (b – a) f(c).

The Mean Value Theorem for integrals is a consequence of the Mean Value Theorem for derivatives and the Fundamental Theorem of Calculus.

The geometric interpretation of the Mean Value Theorem for integrals is that, for positive functions f, there is a number c such that the rectangle with base [a, b] and height f(c) has the same area as the region under the graph f from a to b.

Mean Value Theorem for Integrals

Example : Verify mean value theorem for integrals for the function f(x) = x2 – 2x + 3 on the interval [0 , 2 ].

Solution: We have f(x) = x2 – 2x + 3 on the interval [-1 , 3 ].
Being a polynomial function it is continuous in the given interval.

According to mean value theorem for integrals there is a number c in [ -1,3 ] such that,
$\int_{-1}^{3}(x^{2}-2x+3)dx$ = (3+1)f(c)

⇒  ${f}(c)$ = $\frac{1}{4}$ $\int_{-1}^{3}(x^{2}-2x+3)dx$

               = $\frac{1}{4}$ [$\frac{x^{3}}{3}$ - 2 $\frac{x^{2}}{2}$+3x]                [where the limit vary from -1 to 3]
 
               = $\frac{1}{4}$ [9-9+9-$(\frac{-1}{3}-1-3)$]

               = $\frac{1}{4}$ [9+ $\frac{1}{3}$ + 4] = $\frac{40}{12}$ = $\frac{10}{3}$

     ${f}(c)$ = ($c^{2}$-2c+3) = $\frac{10}{3}$

⇒           ($c^{2}$-2c+3)-$\frac{10}{3}$=0

⇒           (3$c^{2}$-6c-1)=0

⇒           c = 2.15, -0.15

⇒        The two values 2.15 and -0.15 lies in the interval [ -1, 3 ].

Mean Value Theorem Problems


Solved Examples

Question 1: Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. f(x) = $\sqrt{x}$. Find the values of c on the interval [0, 9] 
Solution:
 
Given f(x) = vx on [0,9]
f(9) = v9 = 3
f(0) = v0 = 0

Using mean value theorem
f'(c) = $\frac{f(b) \ - \ f(a)}{b-a}$

f'(c) = $\frac{f(9) \ - \ f(0)}{9 \ - \ 0}$ = $\frac{3-0}{9}$ = $\frac{1}{3}$

f'(x) = $\frac{1}{2}$ vx

f'(c) = $\frac{1}{2}$ vc


$\frac{1}{2}$
vc = $\frac{1}{3}$

2vc = 3
vc = $\frac{3}{2}$

c = $\frac{9}{4}$
 

Question 2: Determine if the given function satisfies the Mean Value Theorem on [-6,6]. If so, find all numbers c on the interval that satisfy the theorem. f(x)= $x^{3}$ - 18 x
Solution:
 
Given f(x)= $x^{3}$ - 18 x

The mean value theorem states that if f is a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a < b, then there exists some c in (a, b) such that f'(c) = $\frac{f(b)-f(a)}{b-a}$

1) f is continuous on [-6,6]
2) f is differentiable on (-6,6)
3) a < b
 
f(6) = $6^{3}$ - 18 $\times$ 6 = 108
f(-6) = $(-6)^{3}$ - 18 $\times$  (-6)= -108

Using mean value theorem
f'(c) = $\frac{f(b)-f(a)}{b-a}$

f'(x) = $\frac{108-(-108)}{6 -( -6)}$

= 18

f'(x) = 3x2 - 18
3x2 - 18 =18
3x2 = 36
x2 = 12
x = $\pm $ 2v3

Hence c = $\pm $ 2v3