PROBLEMS ON LAGRANGES MEAN VALUE THEOREM

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,

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.

Problem 1 :

A race car driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours.

Solution :

Here he is travelling in the interval [0, 2].

f'(x)  =  150 km/hr

f(a)  =  20, f(b)  =  ?

f'(x)  =  [f(b) - f(a)]/(b-a)

150  =  [f(b) - 20]/(2-0)

150(2)  =  f(b) - 20

300+20  =  f(b)

f(b)  =  320

So, he can cover the maximum distance of 320 km/hr.

Problem 2 :

Suppose that for a function f (x), 

f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) − f(1) ≤ 3 .

Solution :

Given :

f'(x) ≤ 1

f'(x)  =  [f(b) - f(a)]/(b-a)

[f(4)-f(1)]/(4-1) ≤ 1

[f(4)-f(1)]/3 ≤ 1

[f(4)-f(1)] ≤ 3

Problem 2 :

Does there exist a differentiable function f(x) such that f(0) = −1, f(2) = 4 and f'(x)  ≤ 2 for all x . Justify your answer.

Solution :

f'(x)  =  [f(b) - f(a)]/(b-a)

Given :

f'(x)  ≤ 2

=  [4-(-1)] / (2-0)

=  5/2

=  2.5

f'(x)  =  2.5, it is not possible, because it is not in the interval [0, 2].

Problem 3 :

Show that there lies a point on the curve

f(x)  =  x(x+3)e^π/2, -3 ≤  x ≤ 0,

where the tangent is parallel to the x-axis.

Solution :

Since the tangent is parallel to x-axis, its slope must be 0.

f'(x)  =  e^π/2[x(1) + (x+3)(1)]

f'(x)  =  e^π/2[2x+3]

e^π/2[2x+3]  =  0

2x+3  =  0

2x  =  -3

x  =  -3/2,  ∊ [-3, 0]

Problem 4 :

Using mean value theorem prove that for a > 0, b > 0,

|e^-a-e^-b| < |a-b|

Solution :

Let f(x)  =  e^-x

f'(x)  =  e^-x

f(a)  =  e^-a

f(b)  =  e^-b

f'(x)  =  [f(b) - f(a)] / (b-a)

|[e^-b - e^-a] / (b-a) | ≤ |e^-c| ≤ 1

|[e^-b - e^-a]| ≤ |b-a|

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