Mean Value Theorem

MEAN‒VALUE THEOREM

(1) First form.

If (i) f(x) is continuous in the closed interval [a, b], and

(ii) f '(x) exists in the open interval (a, b), then there is at least one value c of x in (a, b), such that

f(b)−f(a) / b‒a = f '(c).

Consider the function ϕ(x) = f(x) ‒ [ f(b)‒f(a) / b‒a ] x

Since f(x) is continuous in [a, b];  ∴ ϕ is also continuous in [a, b].

Since f '(x) exists in (a, b);

∴ ϕ'(x) also exists in (a, b) and = f '(x) ‒ [f(b)‒f(a) / b‒a ]   ……(1)

Clearly ϕ(a) = bf(a)‒af(b) / b‒a = ϕ(b).

Thus ϕ(x) satisfies all the conditions of Rolle's theorem.

∴ There at least one value c of x between a and b such that ϕ'(c) = 0. Substituting

 x = c in (1), we get

 ƒ'(c) – [ f(b)−f(a) / b‒a]                     ………(2)

which proves the theorem.

 

(2) Second form.

If we write b = a + h then since a < c < b,

 c = a+ θh where 0 < θ < 1.

Thus the mean value theorem may be stated as follows:

If (i) f(x) is continuous in the closed interval [a, a + h]

and (ii) f '(x) exists in the open interval (a, a + h)

then there is at least one number θ (0 < θ < 1) such that

 f(a + h) = f(a) + hf '(a + θh)

 

(3) Geometrical Interpretation.

Let A, B be the points on the curve y = f(x) corresponding to x=a and x = b so that A = [a, f(a)] and B = [b, f(b)]

∴ Slope of chord AB = f(b)‒f(a) / b‒a

By (2), the slope of the chord AB = f '(c) the slope of the tangent to the curve at c (x = c).

Hence the Lagrange's mean value theorem asserts that if a curve AB has a tangent et each of its points, then there exists at least one point C on this curve, the tangent at which is parallel to the chord AB.

Cor. If f '(x) = 0 in the interval (a, b) then f(x) is constant in [a, b].

For, if x₁, x₂ be any two values of x in (a, b), then by (2),

f(x₂)‒ f(x₁) = (x₂‒x₁) f '(c) = 0(x₁ < c < x₂)

Thus f(x₁) = f(x₂)

i.e. f(x) has the same value every value of x in (a, b).

 

Example 148. Verify the mean value theorem on the given interval, and find all values of c in that interval that satisfy the theorem.

(a) f(x) = x²−x, [‒3,5]

(b) f(x) = √[25 − x²], [‒5,3]

Solution:

(a) The given function

 f(x) = x²‒ x is a polynomial. It is continuous on [‒3,5] and differentiable on (‒3,5). Here a = −3,b = 5.

f(a) = f (‒3) = (‒3)²+3 = 12

f(b) = ƒ(5) = 5² ‒ 5 = 20

f '(x)=2x‒1

f '(c) = 2c ‒ 1

The Lagrange's MVT is

f(b)‒f(a) = (b − a)ƒ'(c)

8 = 8[2c ‒ 1]

2c‒1=1

c=1

Therefore, c is in [‒3,5].

Hence, Mean Value Theorem is verified.

The given function f(x) = √[25‒x²] is continuous on [‒5.3] and differentiable on (−5,3). Here a = −5, b = 3.

f(a) = ƒ (‒5) = √[25−25] = 0

f(b) = f(3) = √[25−9] = 4

f '(x) = [ 1 / 2√(25‒x²)] (‒2x) =   ‒x / √(25‒x²)

Then

f '(c) = ‒c / √(25‒c²)               ……(2)

The Lagrange's MVT is

 f(b)‒f(a) = (b‒a) f '(c)

4 = 8 [ ‒c / √(25‒c²) ]

‒c / √(25‒c²) = ½

√(25‒c²) = ‒2c

Squaring, we get

25‒c² = 4c²

5c² = 25

c² = 5

c = ±√5

 c = ±2.24

Therefore, c is in [‒5,3].

Hence, Mean Value Theorem is verified.

 

Example 149. Verify Lagrange's mean value theorem for the following functions:

a) f(x) = x² + 3x + 2 in 1 ≤ x ≤2

b) f(x) = x³ + x − 1 in [0,2]

c) f(x) = 1/x in −1 ≤ x ≤ 1

d) f(x) = 1 + x^2/3 in [‒8,1]

e) f(x) = x + 1/x in [1/2,2]

f) f(x) = e^−2x in [0,3]

Solution:

a) Given f(x) = x² + 3x + 2 in 1 ≤ x ≤ 2.

Which is a polynomial. It is continuous on [1,2] and differentiable on (1,2).

Here a = 1, b = 2

 f '(x)= 2x + 3       i.e., f ' (c) = 2c + 3

Also f(b) = f(2) = 2² +3(2)+2 = 4+6+2 = 12

f(a) = f(1) = 1² + 3(1) + 2 = 6

The Lagrange's MVT is

f(b)‒f(a) = (b − a) f '(c)

= 12‒6 = (2‒1) (2c+3)

⇒ 6=2c+3

⇒2c = 3

⇒ c = 3/2

i.e., 1 < 3/2 < 2

Hence, Lagrange's MVT is verified for this function.

b) Given ƒ(x) = x³ + x ‒ 1 in [0,2]. Which is a polynomial. It is continuous on [0,2] and differentiable on (0,2).

Here, a = 0, b = 2

 f '(x) = 3x² + 1       i.e, f '(c) = 3c² + 1

Also f(b) = f(2) = 2³ + 2 −1 = 8+2−1 = 9

f(a) = f(0) = 0 + 0 ‒1 = ‒1

The Lagrange's MVT is

f(b) ‒ f(a) = (b − a) f '(c)

⇒ 9 − (−1) = (2‒0)(3c² + 1)

⇒ 10 = 2(3c² + 1)

⇒ 3c²+1 = 5

⇒ 3c² = 4

⇒ c² = 4/3

⇒ c = ± 2/√3 = ±1.1547

ie., 0 < 1.1547 < 2

Hence, the mean value theorem is verified

c) Given f(x) = 1/x in ‒ 1 ≤ x ≤ 1

Here, a = ‒1, b = 1

 f '(x) = ‒1/x²

It is easy to observe that both the functions f(x) and f '(x) do not exist at

 x = 0.

Thus, f(x) is neither continuous nor differentiable on ‒ 1 ≤ x ≤ 1. Therefore, the Lagrange's MVT is not applicable for this function.

d) Given f(x) = 1 + x^2/3 in [‒8,1]

The function f(x) is continuous on [‒8, 1] as the function is finite for all values of x on [‒8,1].

But, f '(x) = 2/3 x^−1/3 does not exist at x = 0

i.e., f(x) is not differentiable in (‒8, 1)

Hence, Lagrange's MVT is not applicable for this function.

e) Given f(x) = x + 1/x, [1/2,2]

Here, a = 1/2, b = 2

 f '(x) = 1 ‒ 1/x²

 f '(c) = 1 ‒ 1/c²

Also f(b) = f(2) = 2 + 1/2 = 5/2

f(a) = f(1/2) = 1/2 + 2 = 5/2

The Lagrange's MVT is

f(b) − f(a) = (b − a)ƒ'(c)

⇒ 5/2 ‒ 5/2 = (2 ‒ 1/2)[ 1‒ 1/c² ]

⇒ 0 = (3/2) [1 ‒ 1/c²]

⇒1 ‒ 1/c² = 0

⇒ c² = 1

⇒ c = ±1

le, 1/2 < c < 1

le, 1/2 < 1 < 2

Hence, Lagrange's MVT is verified for this function.

f) Given f(x) = e^‒2x, [0,3]. It is continuous on [0,3] and differentiable on (0,3).

Here, a = 0, b = 3

f '(x) = (‒2)e^‒2x

f '(c) = (‒2)e^‒2c

Also f(b) = ƒ(3) = e⁻⁶

ƒ(a) = f(0) = e⁻⁰ = 1

The Lagrange's MVT is

f(b)‒f(a) = (b − a)ƒ'(c)

⇒ e^‒6‒1 = (3‒0)(‒2)e^‒2c

⇒e^‒6‒1 = ‒6e^‒2c

⇒e^‒2c = ‒1/6 [ e^‒6 ‒ 1 ]

⇒e^‒2c = ‒1/6 e^‒6 + 1/6

= 1/6[1 + e^‒6]

⇒ loge^‒2c = log [1/6 (1‒e^‒6)]

⇒ ‒2c = log[1/6 (1‒e^‒6) ]

⇒ c = ‒1/2 × log[1/6 (1‒e^‒6) ]

i.e, 0 < c < 3,

Since c = 0.3896

Hence, Lagrange's MVT is verified.

 

Example 150. Suppose that ƒ(0) = − 3 and f '(x) ≤ 5 for all values of x. How large can ƒ(2) possibly be?

Solution:

Given f is differentiable (and therefore continuous) everywhere.

In particular, we can apply the Mean Value Theorem on the interval [0,2].

Lagrange's MVT is

f(b) ‒ f(a) = (b ‒ a) f '(c)

f(2) ‒ f(0) = f '(c)(2 – 0)

ƒ(2) = ƒ(0) + 2ƒ'(c) ... (1)

Given ƒ(0) = ‒3.

(1) ⇒ ƒ(2) = −3+ 2f '(c)            ... (2)

Also given

 f '(x) ≤ 5 for all x, so f '(c) ≤ 5.

We have 2f '(c) ≤ 10,

(2) ⇒ ƒ(2) = − 3 + 2f '(c) ≤ − 3 + 10 = 7

The largest possible value for f(2) is 7.

 

Example 151. In the Mean value theorem, f(b) − f(a) = (b − a)f '(c), determine e lying between a and b, if ƒ(x) = x(x − 1)(x − 2), a = 0 and b = 1/2.

Solution:

 f(a) = 0 ƒ(b) = 1/2 (‒ 1/2) ( − 3/2) = 3/8

ƒ'(x) = 3x² − 6x + 2, f '(c) = 3c² − 6c + 2

Substituting in (i),

3/8 – 0 = (1/2 – 0)(3c² – 6c + 2)

⇒ 12c² ‒ 24c + 5 = 0

 c = { 24±√[(24)²‒12 × 5 × 4] } / 24 = 1±0.764 = 1.764; 0.236.

Since c lies between 0 and 1/2,

c = 0.236

 

Example 152. If f(x) = x³‒ x, a = 0, b = 2, by mean value theorem find the value of c.

Solution:

Since f is a polynomial, it is continuous and differentiable for all x, so it is certainly continuous on [0,2] and differentiable on (0,2). Therefore, by the Mean Value Theorem, there is a number c in (0, 2) such that

f(2) ‒ f(0) = f '(c)(2 – 0)

Now f(2) = 6, f(0) = 0 and f '(x) = 3x²‒1 so this equation becomes

6 = (3c² ‒ 1) 2 = 6c²‒2

which gives c² = 4/3 that c = +2/√3. But c must lie in (0,2), so c = 2√3.

 

 

EXERCISE

 

44. Verify Lagrange's mean value theorem for the following functions:

a) f(x) = x^2/3 in 0≤x≤1

b) f(x) = x³ ‒ x² in [‒1,2]

c) f(x) = x² + 2x ‒ 1 in 0≤ x ≤1

d) f(x) = 2x² ‒ 3x + 1 in [0,2]

e) f(x)=√[x‒1], in [1,3]

f) f (x) = x / x+2 in [1,4]

 

45. Suppose that 3 ≤ f '(x) ≤ 5, for all values of x. Show that 18 ≤ f(8) ‒ f(2) ≤ 30.

 

46. Using Lagrange's Mean Value Theorem prove that

x/(1+x) < log(1+x) < x for all x < 0