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Sagot :
To address the problem, we'll apply Rolle's Theorem to the function [tex]\( f(x) = 2x^2 - 16x - 1 \)[/tex] over the interval [tex]\([2, 6]\)[/tex]. Let's work through it step by step.
### Step 1: Verify the Conditions for Rolle's Theorem
Rolle's Theorem states that if a function [tex]\( f \)[/tex] is:
1. Continuous on a closed interval [tex]\([a, b]\)[/tex],
2. Differentiable on the open interval [tex]\((a, b)\)[/tex],
3. And [tex]\( f(a) = f(b) \)[/tex],
then there exists at least one [tex]\( c \)[/tex] in the interval [tex]\((a, b)\)[/tex] such that [tex]\( f'(c) = 0 \)[/tex].
First, we check that [tex]\( f(x) \)[/tex] is a polynomial, which is continuous and differentiable everywhere. Thus, conditions 1 and 2 are satisfied.
Next, we check if [tex]\( f(2) = f(6) \)[/tex]:
[tex]\[ f(2) = 2(2)^2 - 16(2) - 1 = 8 - 32 - 1 = -25 \][/tex]
[tex]\[ f(6) = 2(6)^2 - 16(6) - 1 = 72 - 96 - 1 = -25 \][/tex]
Since [tex]\( f(2) = f(6) = -25 \)[/tex], condition 3 is also met. Therefore, all conditions for Rolle's Theorem are satisfied.
### Step 2: Find the Derivative of [tex]\( f(x) \)[/tex]
Now, we need to find the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (2x^2 - 16x - 1) = 4x - 16 \][/tex]
### Step 3: Find the Critical Points
Set the derivative equal to zero to find the critical points:
[tex]\[ 4x - 16 = 0 \][/tex]
[tex]\[ 4x = 16 \][/tex]
[tex]\[ x = 4 \][/tex]
### Step 4: Verify the Critical Point is Within the Interval
The critical point [tex]\( x = 4 \)[/tex] lies within the interval [tex]\([2, 6]\)[/tex]. Therefore, this is a valid value for [tex]\( c \)[/tex].
### Conclusion
There is exactly one value of [tex]\( c \)[/tex] such that [tex]\( f'(c) = 0 \)[/tex] within the interval [tex]\([2, 6]\)[/tex], and the value of [tex]\( c \)[/tex] is [tex]\( 4 \)[/tex].
Summary:
- Number of [tex]\( c \)[/tex] values: [tex]\( \boxed{1} \)[/tex]
- Value of [tex]\( c \)[/tex]: [tex]\( \boxed{4} \)[/tex]
### Step 1: Verify the Conditions for Rolle's Theorem
Rolle's Theorem states that if a function [tex]\( f \)[/tex] is:
1. Continuous on a closed interval [tex]\([a, b]\)[/tex],
2. Differentiable on the open interval [tex]\((a, b)\)[/tex],
3. And [tex]\( f(a) = f(b) \)[/tex],
then there exists at least one [tex]\( c \)[/tex] in the interval [tex]\((a, b)\)[/tex] such that [tex]\( f'(c) = 0 \)[/tex].
First, we check that [tex]\( f(x) \)[/tex] is a polynomial, which is continuous and differentiable everywhere. Thus, conditions 1 and 2 are satisfied.
Next, we check if [tex]\( f(2) = f(6) \)[/tex]:
[tex]\[ f(2) = 2(2)^2 - 16(2) - 1 = 8 - 32 - 1 = -25 \][/tex]
[tex]\[ f(6) = 2(6)^2 - 16(6) - 1 = 72 - 96 - 1 = -25 \][/tex]
Since [tex]\( f(2) = f(6) = -25 \)[/tex], condition 3 is also met. Therefore, all conditions for Rolle's Theorem are satisfied.
### Step 2: Find the Derivative of [tex]\( f(x) \)[/tex]
Now, we need to find the derivative [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} (2x^2 - 16x - 1) = 4x - 16 \][/tex]
### Step 3: Find the Critical Points
Set the derivative equal to zero to find the critical points:
[tex]\[ 4x - 16 = 0 \][/tex]
[tex]\[ 4x = 16 \][/tex]
[tex]\[ x = 4 \][/tex]
### Step 4: Verify the Critical Point is Within the Interval
The critical point [tex]\( x = 4 \)[/tex] lies within the interval [tex]\([2, 6]\)[/tex]. Therefore, this is a valid value for [tex]\( c \)[/tex].
### Conclusion
There is exactly one value of [tex]\( c \)[/tex] such that [tex]\( f'(c) = 0 \)[/tex] within the interval [tex]\([2, 6]\)[/tex], and the value of [tex]\( c \)[/tex] is [tex]\( 4 \)[/tex].
Summary:
- Number of [tex]\( c \)[/tex] values: [tex]\( \boxed{1} \)[/tex]
- Value of [tex]\( c \)[/tex]: [tex]\( \boxed{4} \)[/tex]
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