Verify Rolle's theorem for the function f(x) = x2 + 4 in [–3, 3].

1.	Verify Rolle's theorem for the function f(x) = x2 + 4 in [–3, 3].
Answer» Given f(x) = x² + 4 Since f is a second degree polynomial ∴ f is continuous on [–3, 3] and f is desivable on (–3, 3) Also f(–3) = 9 + 4 = 13 f(3) = 9 + 4 = 13 ∴ f(–3) = f(3) ∴ f satisfies all the conditions of Role's theorum. ∴ There exists c∈(–3, 3) such that f¹(c) = 0 But f¹(x) = 2x f¹(c) = 2c 0 = 2c ⇒ c = 0 ∈ (–3, 3) Hence Rolle's theorum is verified.

Answer»

Given f(x) = x² + 4

Since f is a second degree polynomial

∴ f is continuous on [–3, 3] and f is desivable on (–3, 3)

Also f(–3) = 9 + 4 = 13 f(3) = 9 + 4 = 13

∴ f(–3) = f(3)

∴ f satisfies all the conditions of Role's theorum.

∴ There exists c∈(–3, 3) such that f¹(c) = 0

But f¹(x) = 2x

f¹(c) = 2c

0 = 2c

⇒ c = 0 ∈ (–3, 3)

Hence Rolle's theorum is verified.

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Verify Rolle's theorem for the function f(x) = x2 + 4 in [–3, 3].

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