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Sagot :
Let's verify the hypotheses of Rolle's Theorem and then find the values of [tex]\( c \)[/tex] that satisfy its conclusion for the function [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] on the interval [tex]\([2, 4]\)[/tex].
### Hypotheses of Rolle's Theorem:
1. Continuity on the closed interval [tex]\([2, 4]\)[/tex]:
- The function [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] is a polynomial.
- Polynomial functions are continuous for all real numbers.
- Hence, [tex]\( f(x) \)[/tex] is continuous on [tex]\([2, 4]\)[/tex].
2. Differentiability on the open interval [tex]\((2, 4)\)[/tex]:
- The function [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] is a polynomial.
- Polynomial functions are differentiable for all real numbers.
- Hence, [tex]\( f(x) \)[/tex] is differentiable on [tex]\((2, 4)\)[/tex].
3. [tex]\( f(a) = f(b) \)[/tex] where [tex]\( a = 2 \)[/tex] and [tex]\( b = 4 \)[/tex]:
- Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5 - 12(2) + 2(2)^2 = 5 - 24 + 8 = -11. \][/tex]
- Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 5 - 12(4) + 2(4)^2 = 5 - 48 + 32 = -11. \][/tex]
- We have [tex]\( f(2) = f(4) = -11 \)[/tex], so [tex]\( f(2) = f(4) \)[/tex] is satisfied.
Since all three conditions of Rolle's Theorem are satisfied, there exists at least one number [tex]\( c \)[/tex] in the open interval [tex]\((2, 4)\)[/tex] such that [tex]\( f'(c) = 0 \)[/tex].
### Finding the value(s) of [tex]\( c \)[/tex]:
1. Find the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}[5 - 12x + 2x^2] = -12 + 4x. \][/tex]
2. Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ f'(c) = -12 + 4c = 0 \][/tex]
[tex]\[ 4c = 12 \][/tex]
[tex]\[ c = 3. \][/tex]
3. Verify that [tex]\( c = 3 \)[/tex] is in the open interval [tex]\((2, 4)\)[/tex]:
- Since [tex]\( 3 \)[/tex] is between [tex]\( 2 \)[/tex] and [tex]\( 4 \)[/tex], [tex]\( c = 3 \)[/tex] is a valid solution.
Therefore, the number [tex]\( c \)[/tex] that satisfies Rolle's Theorem for [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] on the interval [tex]\([2, 4]\)[/tex] is:
[tex]\[ c = 3 \][/tex]
Thus, the final answer is:
[tex]\[ c = 3 \][/tex]
### Hypotheses of Rolle's Theorem:
1. Continuity on the closed interval [tex]\([2, 4]\)[/tex]:
- The function [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] is a polynomial.
- Polynomial functions are continuous for all real numbers.
- Hence, [tex]\( f(x) \)[/tex] is continuous on [tex]\([2, 4]\)[/tex].
2. Differentiability on the open interval [tex]\((2, 4)\)[/tex]:
- The function [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] is a polynomial.
- Polynomial functions are differentiable for all real numbers.
- Hence, [tex]\( f(x) \)[/tex] is differentiable on [tex]\((2, 4)\)[/tex].
3. [tex]\( f(a) = f(b) \)[/tex] where [tex]\( a = 2 \)[/tex] and [tex]\( b = 4 \)[/tex]:
- Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5 - 12(2) + 2(2)^2 = 5 - 24 + 8 = -11. \][/tex]
- Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 5 - 12(4) + 2(4)^2 = 5 - 48 + 32 = -11. \][/tex]
- We have [tex]\( f(2) = f(4) = -11 \)[/tex], so [tex]\( f(2) = f(4) \)[/tex] is satisfied.
Since all three conditions of Rolle's Theorem are satisfied, there exists at least one number [tex]\( c \)[/tex] in the open interval [tex]\((2, 4)\)[/tex] such that [tex]\( f'(c) = 0 \)[/tex].
### Finding the value(s) of [tex]\( c \)[/tex]:
1. Find the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}[5 - 12x + 2x^2] = -12 + 4x. \][/tex]
2. Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ f'(c) = -12 + 4c = 0 \][/tex]
[tex]\[ 4c = 12 \][/tex]
[tex]\[ c = 3. \][/tex]
3. Verify that [tex]\( c = 3 \)[/tex] is in the open interval [tex]\((2, 4)\)[/tex]:
- Since [tex]\( 3 \)[/tex] is between [tex]\( 2 \)[/tex] and [tex]\( 4 \)[/tex], [tex]\( c = 3 \)[/tex] is a valid solution.
Therefore, the number [tex]\( c \)[/tex] that satisfies Rolle's Theorem for [tex]\( f(x) = 5 - 12x + 2x^2 \)[/tex] on the interval [tex]\([2, 4]\)[/tex] is:
[tex]\[ c = 3 \][/tex]
Thus, the final answer is:
[tex]\[ c = 3 \][/tex]
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