text: # For what values of x is f(x)= x^2-x + 1/x-1/x^2  concave or convex?
  
  ##### 1 Answer
  Nov 19, 2017
  
  $x = - 1 , 3 , \frac{3 \pm \sqrt{21}}{2}$
  
  #### Explanation:
  
  Find the second derivative of $f \left(x\right)$.
  $f ' \left(x\right) = 2 x - 1 - \frac{1}{x} ^ 2 + \frac{2}{x} ^ 3$
  $f ' ' \left(x\right) = 2 + \frac{2}{x} ^ 3 - \frac{6}{x} ^ 4$
  
  Find the critical points of $f ' ' \left(x\right)$.
  $0 = 2 + \frac{2}{x} ^ 3 - \frac{6}{x} ^ 4$
  $0 = 2 \left(1 + \frac{1}{x} ^ 3 - \frac{3}{x} ^ 4\right)$
  $- 1 = \frac{1}{x} ^ 3 - \frac{3}{x} ^ 4$
  $- 1 = {x}^{3} - {x}^{4} / 3$
  $- 1 = x \left({x}^{2} - {x}^{3} / 3\right)$
  
  $x = - 1$
  
  $- 1 = {x}^{2} - {x}^{3} / 3$
  ${x}^{3} / 3 - 1 = {x}^{2}$
  $\frac{{x}^{3} - 3}{3} = {x}^{2}$
  ${x}^{3} - 3 = 3 {x}^{2}$
  ${x}^{3} - 3 {x}^{2} = 3$
  $x \left({x}^{2} - 3 x\right) = 3$
  
  $x = 3$
  
  ${x}^{2} - 3 x = 3$
  $0 = {x}^{2} - 3 x - 3$
  
  Use the quadratic formula.
  
  $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{3 \pm \sqrt{21}}{2}$
  
  $x = - 1 , 3 , \frac{3 \pm \sqrt{21}}{2}$
explanation: The document contains a math question asking for the values of \( x \) for which the function \( f(x) = x^2 - x + \frac{1}{x} - \frac{1}{x^2} \) is concave or convex. The document also includes a detailed answer that involves finding the second derivative of the function and solving for the critical points.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # For what values of x is f(x)= x^2-x + 1/x-1/x^2  concave or convex?
  
  ##### 1 Answer
  Nov 19, 2017
  
  $x = - 1 , 3 , \frac{3 \pm \sqrt{21}}{2}$
  
  #### Explanation:
  
  Find the second derivative of $f \left(x\right)$.
  $f ' \left(x\right) = 2 x - 1 - \frac{1}{x} ^ 2 + \frac{2}{x} ^ 3$
  $f ' ' \left(x\right) = 2 + \frac{2}{x} ^ 3 - \frac{6}{x} ^ 4$
  
  Find the critical points of $f ' ' \left(x\right)$.
  $0 = 2 + \frac{2}{x} ^ 3 - \frac{6}{x} ^ 4$
  $0 = 2 \left(1 + \frac{1}{x} ^ 3 - \frac{3}{x} ^ 4\right)$
  $- 1 = \frac{1}{x} ^ 3 - \frac{3}{x} ^ 4$
  $- 1 = {x}^{3} - {x}^{4} / 3$
  $- 1 = x \left({x}^{2} - {x}^{3} / 3\right)$
  
  $x = - 1$
  
  $- 1 = {x}^{2} - {x}^{3} / 3$
  ${x}^{3} / 3 - 1 = {x}^{2}$
  $\frac{{x}^{3} - 3}{3} = {x}^{2}$
  ${x}^{3} - 3 = 3 {x}^{2}$
  ${x}^{3} - 3 {x}^{2} = 3$
  $x \left({x}^{2} - 3 x\right) = 3$
  
  $x = 3$
  
  ${x}^{2} - 3 x = 3$
  $0 = {x}^{2} - 3 x - 3$
  
  Use the quadratic formula.
  
  $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{3 \pm \sqrt{21}}{2}$
  
  $x = - 1 , 3 , \frac{3 \pm \sqrt{21}}{2}$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question asking for the values of \( x \) for which the function \( f(x) = x^2 - x + \frac{1}{x} - \frac{1}{x^2} \) is concave or convex. The document also includes a detailed answer that involves finding the second derivative of the function and solving for the critical points.
  
  Label: question-and-answer