Q. 20
Verify Rolle’s theorem for each of the following functions:
Answer :
Condition (1):
Since, f(x)=x(x+2)ex is a combination of exponential and polynomial function which is continuous for all xϵR.
⇒ f(x)= x(x+2)ex is continuous on [-2,0].
Condition (2):
Here, f’(x)=(x2+4x+2)ex which exist in [-2,0].
So, f(x)=x(x+2)ex is differentiable on (-2,0).
Condition (3):
Here, f(-2)= (-2)(-2+2)e-2 =0
And f(0)= 0(0+2)e0=0
i.e. f(-2)=f(0)
Conditions of Rolle’s theorem are satisfied.
Hence, there exist at least one cϵ(-2,0) such that f’(c)=0
i.e. (c2+4c+2)ec =0
i.e. (c+√2)2=0
i.e. c=-√2
Value of c=-√2 ϵ(-2,0)
Thus, Rolle’s theorem is satisfied.
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