Given: f(x) = x³ – 2x² – x + 3 in [0,1]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

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

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

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

Condition 1:

f(x) = x³ – 2x² – x + 3

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R

⇒ f(x) = x³ – 2x² – x + 3 is continuous at x ∈ [0,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = x³ – 2x² – x + 3

Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R

f’(x) = 3x² – 4x – 1

⇒ f(x) is differentiable at [0,1]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function

Now,

f(x) = x³ – 2x² – x + 3 x ∈ [0,1]

f(a) = f(0) = 3

f(b) = f(1) = (1)³ – 2(1)² – 1 + 3

= 1 – 2 – 1 + 3

= 4 – 3

= 1

Now, let us show that there exist c ∈ (0,1) such that

f(x) = x³ – 2x² – x + 3

On differentiating above with respect to x, we get

f’(x) = 3x² – 4x – 1

Put x = c in above equation, we get

f’(c) = 3c² – 4c – 1 …(i)

By Mean Value Theorem,

⇒ f’(c) = -2

⇒ 3c² – 4c – 1 = -2 [from (i)]

⇒ 3c² – 4c -1 + 2 = 0

⇒ 3c² – 4c + 1 = 0

On factorising, we get

⇒ 3c² – 3c – c + 1 = 0

⇒ 3c(c – 1) – 1(c – 1) = 0

⇒ (3c – 1) (c – 1) = 0

⇒ (3c – 1) = 0 or (c – 1) = 0

So, value of

Thus, Mean Value Theorem is verified.