Verify Mean Value Theorem, if f (x) = x2 – 4x – 3 in the interval

1.	Verify Mean Value Theorem, if f (x) = x2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Answer» Here, f(x) = x² − 4x − 3, x ∈ [1, 4] which is a polynomial function, so it is continuous and derivable at all x ∈ R, therefore (i) f(x) is continuous on [1, 4] (ii) f(x) is derivable on (1, 4). Therefore, Conditions of Lagrange's theorem are satisfied on [1, 4]. Hence, there is atleast one real number. c ∈ (1, 4) such that Now, f'(x) = d/dx(x² - 4x - 3) = 2x - 4 ⇒ f(4) = 4² - 4(4) - 3 = 16 - 16 - 3 = - 3 and f(1) = 1 - 4 - 3 = - 6 ∴ f'(c) =(f(4) - f(1))/(4 - 1) = (-3 - (-6))/3 = (-3 + 6)/3 = 1 ⇒ 2c - 4 = 1 ⇒ 2c = 1 + 4 = 5 ⇒ c = 5/2 ∈ (1, 4)

Verify Mean Value Theorem, if f (x) = x2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Answer»

Here, f(x) = x² − 4x − 3, x ∈ [1, 4] which is a polynomial function, so it is continuous and derivable at all x ∈ R, therefore

(i) f(x) is continuous on [1, 4]

(ii) f(x) is derivable on (1, 4).

Therefore, Conditions of Lagrange's theorem are satisfied on [1, 4].

Hence, there is atleast one real number. c ∈ (1, 4) such that

Now, f'(x) = d/dx(x² - 4x - 3) = 2x - 4

⇒ f(4) = 4² - 4(4) - 3 = 16 - 16 - 3 = - 3

and f(1) = 1 - 4 - 3 = - 6

∴ f'(c) =(f(4) - f(1))/(4 - 1) = (-3 - (-6))/3 = (-3 + 6)/3 = 1

⇒ 2c - 4 = 1 ⇒ 2c = 1 + 4 = 5 ⇒ c = 5/2 ∈ (1, 4)

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