WHY ROLLES THEOREM IS NOT APPLICABLE FOR THE GIVEN FUNCTION

Problem 1 :

Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.

(i)  f(x)  =  |(1/x)|, x ∊ [-1, 1]          Solution

(ii) f (x) = tan x, x ∊ [0, π]       Solution

(iii)  f(x)  =  x - 2 log x, x ∊ [2, 7]            Solution

Problem 2 :

Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x -axis for the following functions :

(i) f(x)  =  x² − x, x ∈ [0, 1]            Solution

(ii)  f(x)  =  (x²-2x)/(x+2), x ∈ [-1, 6]        Solution

(iii)  f(x)  =   √x - (x/3), x∈ [0, 9]          Solution

Problem 3 :

Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals :

(i)  f(x)  =  (x+1)/x, x ∊ [-1, 2]      Solution

(ii)  f(x)  =  |3x+1|, x ∊ [-1, 3]      Solution

Problem 4 :

Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:

(i)  f (x)  =  x³ - 3x + 2, x ∊ [-2, 2]             Solution

(ii)  f(x)  =  (x-2)(x-7),  x ∊ [3, 11]          Solution

Problem 5 :

Show that the value in the conclusion of the mean value theorem for

(i) f (x) = 1/x on a closed interval of positive numbers [a, b] is √ab            Solution

(ii) f (x) = Ax² + Bx + C on any interval [a, b] is (a + b)/2

Solution

Problem 6 :

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

Problem 7 :

Suppose that for a function f (x),  f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) − f(1) ≤ 3.          Solution

Problem 8 :

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

Problem 9 :

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

Problem 10 :

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

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

Solution

(1) (i)   f(0)  =  ∞  (ii)  f(π/2)  =  ∞  (ii)  f(2)  ≠ f(7)

(2)  (i)  c  =  1/2  (ii)  x  =  -2±2√2  (iii)  x  =  9/4 

(3)  (i)  Not continuous on [-1, 2].

(ii)  not differentiable at x  =  -1/3

(4)  (i)  x  =  ±2/√3  (ii)  x  =  7

(5)  (i)  x  =  √ab  (ii)  x  =  (a+b)/2

(6)   320 km/hr.

(7)  [f(4)-f(1)] ≤ 3

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

(9)  x  =  -3/2,

(10)  |[e^-b - e^-a]| ≤ |b-a|

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