text: 1. ## rolle's theorem
  
  for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
  
  2. Originally Posted by wisezeta
  for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
  try instead the Intermediate Value Theorem.
  
  3. Originally Posted by skeeter
  try instead the Intermediate Value Theorem.
  can u do it using rolle's theorem?
  
  4. Originally Posted by airportman92
  can u do it using rolle's theorem?
  Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.
  
  I'm not saying it can't be done, but it would not be as simple as using the IVT.
  
  5. Originally Posted by skeeter
  Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.
  
  I'm not saying it can't be done, but it would not be as simple as using the IVT.
  
  how would you do it using rolle's theorem though just for fun
  
  6. Originally Posted by airportman92
  how would you do it using rolle's theorem though just for fun
  it would not be fun.
  
  7. Originally Posted by skeeter
  it would not be fun.
  but how do u do it hah
  
  8. Originally Posted by airportman92
  but how do u do it hah
  I'm not sure.
  
  9. someone how do you use rolles theorem to do this one lol
  
  10. It's a useless detour (at least how I did it is), but since you want it so badly:
  
  Let $g_c(x)= \frac{x^4}{4} + x^3 -2x +c=4x(x^3+4x^2-2)+c$ Let $h(x)=x^3+4x^2-2$ then $h(0)<0$ and $h(2)>0$ so by the intermediate value theorem there exists $d\in (0,2)$ such that $h(d)=0$ and so $g_c(0)=g_c(d)=c$ and by Rolle's theorem there exists $y\in(0,d)$ such that $g'_c (y)=0$ but $g'_c=g$
  
  As skeeter said, not pretty at all.
explanation: The document is a series of forum posts discussing a math problem. The initial post presents a math question about using Rolle's Theorem to show that there is one real root of the function \( g(x) = x^3 + 3x^2 - 2 \) on the interval \([0, 2]\). The subsequent posts include a suggestion to use the Intermediate Value Theorem instead, a discussion on the applicability of Rolle's Theorem, and a detailed, albeit complex, solution using Rolle's Theorem.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: 1. ## rolle's theorem
  
  for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
  
  2. Originally Posted by wisezeta
  for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
  try instead the Intermediate Value Theorem.
  
  3. Originally Posted by skeeter
  try instead the Intermediate Value Theorem.
  can u do it using rolle's theorem?
  
  4. Originally Posted by airportman92
  can u do it using rolle's theorem?
  Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.
  
  I'm not saying it can't be done, but it would not be as simple as using the IVT.
  
  5. Originally Posted by skeeter
  Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.
  
  I'm not saying it can't be done, but it would not be as simple as using the IVT.
  
  how would you do it using rolle's theorem though just for fun
  
  6. Originally Posted by airportman92
  how would you do it using rolle's theorem though just for fun
  it would not be fun.
  
  7. Originally Posted by skeeter
  it would not be fun.
  but how do u do it hah
  
  8. Originally Posted by airportman92
  but how do u do it hah
  I'm not sure.
  
  9. someone how do you use rolles theorem to do this one lol
  
  10. It's a useless detour (at least how I did it is), but since you want it so badly:
  
  Let $g_c(x)= \frac{x^4}{4} + x^3 -2x +c=4x(x^3+4x^2-2)+c$ Let $h(x)=x^3+4x^2-2$ then $h(0)<0$ and $h(2)>0$ so by the intermediate value theorem there exists $d\in (0,2)$ such that $h(d)=0$ and so $g_c(0)=g_c(d)=c$ and by Rolle's theorem there exists $y\in(0,d)$ such that $g'_c (y)=0$ but $g'_c=g$
  
  As skeeter said, not pretty at all.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a series of forum posts discussing a math problem. The initial post presents a math question about using Rolle's Theorem to show that there is one real root of the function \( g(x) = x^3 + 3x^2 - 2 \) on the interval \([0, 2]\). The subsequent posts include a suggestion to use the Intermediate Value Theorem instead, a discussion on the applicability of Rolle's Theorem, and a detailed, albeit complex, solution using Rolle's Theorem.
  
  Label: question-and-answer