Question

1.) Verify Rolle's Theorem For the Function f(x) = x² + 2x - 8 , x∈ [-4,2]​

Answer 1

Answer:

f(x)=x

2

+2x−8

⇒f(x)=(x+4)(x−2)

f

′

(x)=2x+2

We know every polynomial function is continuous and differentiable in R.

In particular f is continuous and differentiable in [−4,2]

Also f(−4)=f(2)=0

Thus there will exist c∈(−4,2) such that f

′

(c)=0

⇒2c+2=0⇒c=−1∈(−4,2)

Hence, Rolle's theorem verified.

Answer 2

Given:

• f(x) = x² + 2x - 8 , x∈ [-4,2]​

Conditions for Rolle's theorem:-

For a function f:[a,b]→R,

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

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

3.f(a) = f(b)

There exist's some c in (a,b) such that f'(c) = 0 , if the above 3 conditions are satisfied.

Now,

Condition-(1):

As f(x) = x² + 2x - 8 is a polynomial , f(x) is continous for all real value of x.

Hence, the function is continous at x∈ [-4,2]​.

Condition-(2):-

As f(x) = x² + 2x - 8 is a polynomial & we know every polynomial function is differentiable for all x∈R.

Hence, the function is differentiable at x∈ [-4,2]​.

Condition-(3):-

Given,

f(x) = x² + 2x - 8

Putting x = -4 & x =2,

→f(-4) = (-4)² +2(-4)-8

        = 16-8-8 = 0

→f(2) = (2)² + 2(2)-8

        = 4 + 4 - 8 = 0

Therefore,

→ f(-4) = f(2)

Now,

⇒ f(x) = x² + 2x - 8

⇒ f'(x) = 2x+ 2 - 0

⇒ f'(x) = 2x + 2

⇒ f'(c) = 2c+2

Hence,

All the conditions are satisfied.

⇒ f'(c) = 0

⇒ 2c + 2 = 0

⇒ 2c = -2

⇒ c = -2/2

⇒ c = -1

Therefore,c = -1 ∈ (-4,2)

Hence,Rolle's theorem is verified!

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