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To determine whether Rolle's Theorem can be applied to the function [tex]\( f \)[/tex] on the closed interval [tex]\([a, b]\)[/tex], we need to verify the following three conditions of Rolle's Theorem:
1. [tex]\( f \)[/tex] is continuous on the closed interval [tex]\([a, b]\)[/tex].
2. [tex]\( f \)[/tex] is differentiable on the open interval [tex]\((a, b)\)[/tex].
3. [tex]\( f(a) = f(b) \)[/tex].
Let's analyze each condition in detail for the function [tex]\( f(x) = x^{2/3} - 8 \)[/tex] on the interval [tex]\([-27, 27]\)[/tex].
### 1. Continuity
The function [tex]\( f(x) = x^{2/3} - 8 \)[/tex] is a composition of elementary functions [tex]\( x^{2/3} \)[/tex] and a constant [tex]\( -8 \)[/tex]. The function [tex]\( x^{2/3} \)[/tex] is continuous everywhere. Consequently, the function [tex]\( f(x) \)[/tex] is continuous on the closed interval [tex]\([−27, 27]\)[/tex].
### 2. Differentiability
To determine if [tex]\( f \)[/tex] is differentiable on the open interval [tex]\((a, b) = (−27, 27)\)[/tex], we take the derivative of [tex]\( f \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^{2/3} - 8) \][/tex]
[tex]\[ f'(x) = \frac{2}{3}x^{-1/3} \][/tex]
The function [tex]\( f'(x) = \frac{2}{3}x^{-1/3} \)[/tex] is not defined at [tex]\(x = 0\)[/tex] because it involves a negative fractional power of [tex]\( x \)[/tex]. This means [tex]\( f \)[/tex] is not differentiable at [tex]\( x = 0 \)[/tex]. Therefore, [tex]\( f \)[/tex] is not differentiable on the entire open interval [tex]\((−27, 27)\)[/tex].
### 3. Equal Function Values at Endpoints
Next, we check if [tex]\( f(a) = f(b) \)[/tex]:
[tex]\[ f(-27) = (-27)^{2/3} - 8 = 9 - 8 = 1 \][/tex]
[tex]\[ f(27) = 27^{2/3} - 8 = 9 - 8 = 1 \][/tex]
Since [tex]\( f(-27) = f(27) \)[/tex], this condition is satisfied.
### Conclusion
Rolle's Theorem cannot be applied because the function [tex]\( f(x) \)[/tex] is not differentiable at [tex]\( x = 0 \)[/tex] within the open interval [tex]\( (-27, 27) \)[/tex].
### Values of [tex]\( c \)[/tex]
Since Rolle's Theorem cannot be applied, we should enter "NA" for the values of [tex]\( c \)[/tex]:
[tex]\[ c = \boxed{\text{NA}} \][/tex]
So summarizing the answers, the appropriate response is:
- No, because [tex]\( f \)[/tex] is not differentiable in the open interval [tex]\( (a, b) \)[/tex].
And for the values of [tex]\( c \)[/tex]:
[tex]\[ c = \boxed{\text{NA}} \][/tex]
1. [tex]\( f \)[/tex] is continuous on the closed interval [tex]\([a, b]\)[/tex].
2. [tex]\( f \)[/tex] is differentiable on the open interval [tex]\((a, b)\)[/tex].
3. [tex]\( f(a) = f(b) \)[/tex].
Let's analyze each condition in detail for the function [tex]\( f(x) = x^{2/3} - 8 \)[/tex] on the interval [tex]\([-27, 27]\)[/tex].
### 1. Continuity
The function [tex]\( f(x) = x^{2/3} - 8 \)[/tex] is a composition of elementary functions [tex]\( x^{2/3} \)[/tex] and a constant [tex]\( -8 \)[/tex]. The function [tex]\( x^{2/3} \)[/tex] is continuous everywhere. Consequently, the function [tex]\( f(x) \)[/tex] is continuous on the closed interval [tex]\([−27, 27]\)[/tex].
### 2. Differentiability
To determine if [tex]\( f \)[/tex] is differentiable on the open interval [tex]\((a, b) = (−27, 27)\)[/tex], we take the derivative of [tex]\( f \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^{2/3} - 8) \][/tex]
[tex]\[ f'(x) = \frac{2}{3}x^{-1/3} \][/tex]
The function [tex]\( f'(x) = \frac{2}{3}x^{-1/3} \)[/tex] is not defined at [tex]\(x = 0\)[/tex] because it involves a negative fractional power of [tex]\( x \)[/tex]. This means [tex]\( f \)[/tex] is not differentiable at [tex]\( x = 0 \)[/tex]. Therefore, [tex]\( f \)[/tex] is not differentiable on the entire open interval [tex]\((−27, 27)\)[/tex].
### 3. Equal Function Values at Endpoints
Next, we check if [tex]\( f(a) = f(b) \)[/tex]:
[tex]\[ f(-27) = (-27)^{2/3} - 8 = 9 - 8 = 1 \][/tex]
[tex]\[ f(27) = 27^{2/3} - 8 = 9 - 8 = 1 \][/tex]
Since [tex]\( f(-27) = f(27) \)[/tex], this condition is satisfied.
### Conclusion
Rolle's Theorem cannot be applied because the function [tex]\( f(x) \)[/tex] is not differentiable at [tex]\( x = 0 \)[/tex] within the open interval [tex]\( (-27, 27) \)[/tex].
### Values of [tex]\( c \)[/tex]
Since Rolle's Theorem cannot be applied, we should enter "NA" for the values of [tex]\( c \)[/tex]:
[tex]\[ c = \boxed{\text{NA}} \][/tex]
So summarizing the answers, the appropriate response is:
- No, because [tex]\( f \)[/tex] is not differentiable in the open interval [tex]\( (a, b) \)[/tex].
And for the values of [tex]\( c \)[/tex]:
[tex]\[ c = \boxed{\text{NA}} \][/tex]
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