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]e^{3/2}

= 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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