Q. 68

Verify the Rolle’s theorem for each of the functions

f(x) = x(x + 3)e-x/2 in [–3, 0].

Answer :

Given: f(x) = x(x + 3)e-x/2


Now, we have to show that f(x) verify the Rolle’s Theorem


First of all, Conditions of Rolle’s theorem are:


a) f(x) is continuous at (a,b)


b) f(x) is derivable at (a,b)


c) f(a) = f(b)


If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0


Condition 1:



Since, f(x) is multiplication of algebra and exponential function and is defined everywhere in its domain.


is continuous at x [-3,0]


Hence, condition 1 is satisfied.


Condition 2:



On differentiating f(x) with respect to x, we get


[by product rule]











f(x) is differentiable at [-3,0]


Hence, condition 2 is satisfied.


Condition 3:




= [9 – 9]e3/2


= 0



= 0


Hence, f(-3) = f(0)


Hence, condition 3 is also satisfied.


Now, let us show that c (0,1) such that f’(c) = 0



On differentiating above with respect to x, we get



Put x = c in above equation, we get



, all the three conditions of Rolle’s theorem are satisfied


f’(c) = 0




⇒ (c – 3)(c + 2) = 0


c – 3 = 0 or c + 2 = 0


c = 3 or c = -2


So, value of c = -2, 3


c = -2 (-3, 0) but c = 3 (-3, 0)


c = -2


Thus, Rolle’s theorem is verified.


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