Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem. f(x)=5 - 12x + 3x², [1, 3]

Solution:

Given, the function is f(x) = 5 - 12x + 3x²

We have to find all numbers that satisfy the conclusion of Rolle’s theorem in the interval [1, 3].

Rolle’s theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

All polynomials are continuous and differentiable on the set of real numbers.

In the given interval, the function becomes

f(1) = 5 - 12(1) + 3(1)²

= 5 - 12 + 3

= 8 - 12

= -4

f(3) = 5 - 12(3) + 3(3)²

= 5 - 36 + 3(9)

= 5 - 36 + 27

= 32 - 36

= -4

This implies that there is c in the interval [1, 3], such that f’(c) = 0.

On differentiating,

f’(c) = 0 - 12 + 6x

f’(c) = 6x - 12

Solving for f’(c) = 0 for c,

6c - 12 = 0

6c = 12

c = 12/6

c = 2

Therefore, the value of c is 2.

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Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem. f(x)=5 - 12x + 3x², [1, 3]

Summary:

On verifying the function satisfies the three hypotheses of Rolle’s Theorem on the given interval, then all numbers c that satisfy the conclusion of Rolle’s Theorem. f(x)=5 - 12x + 3x², [1, 3] is 2.