text: # What are the extrema of f(x)=3x^2 - 12x + 13 on [-oo,oo]?
  
  Oct 17, 2016
  
  $f \left(x\right)$ has a minimum at $x = 2$
  
  #### Explanation:
  
  Before proceeding, note that this is an upwards facing parabola, meaning we can know without further calculation that it will have no maxima, and a single minimum at its vertex. Completing the square would show us that $f \left(x\right) = 3 {\left(x - 2\right)}^{2} + 1$, giving the vertex, and thus the sole minimum, at $x = 2$. Let's see how this would be done with calculus, though.
  
  Any extrema will occur either at a critical point or at an endpoint of the given interval. As our given interval of $\left(- \infty , \infty\right)$ is open, we can ignore the possibility of endpoints, and so we will first identify the critical points of the function, that is, the point at which the derivative of the function is $0$ or does not exist.
  
  $f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 12 x + 13\right) = 6 x - 12$
  
  Setting this equal to $0$, we find a critical point at $x = 2$
  
  $6 x - 12 = 0 \implies x = \frac{12}{6} = 2$
  
  Now, we can either test to see whether it is an extremum (and what type) by checking some values of $f$ around that point, or by using the second derivative test. Let's use the latter.
  
  $\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left(6 x - 12\right) = 6$
  
  As $f ' ' \left(2\right) = 6 > 0$, the second derivative test tells us that $f \left(x\right)$ has a local minimum at $x = 2$
  
  Thus, using $f ' \left(x\right)$ and $f ' ' \left(x\right)$, we find that $f \left(x\right)$ has a minimum at $x = 2$, matching the result we found using algebra.
explanation: The document contains a math question and a detailed answer. The question asks about the extrema of the function \( f(x) = 3x^2 - 12x + 13 \) on the interval \((- \infty, \infty)\). The answer provides a step-by-step solution using calculus, including finding the critical points and using the second derivative test to confirm the nature of the extrema.
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: # What are the extrema of f(x)=3x^2 - 12x + 13 on [-oo,oo]?
  
  Oct 17, 2016
  
  $f \left(x\right)$ has a minimum at $x = 2$
  
  #### Explanation:
  
  Before proceeding, note that this is an upwards facing parabola, meaning we can know without further calculation that it will have no maxima, and a single minimum at its vertex. Completing the square would show us that $f \left(x\right) = 3 {\left(x - 2\right)}^{2} + 1$, giving the vertex, and thus the sole minimum, at $x = 2$. Let's see how this would be done with calculus, though.
  
  Any extrema will occur either at a critical point or at an endpoint of the given interval. As our given interval of $\left(- \infty , \infty\right)$ is open, we can ignore the possibility of endpoints, and so we will first identify the critical points of the function, that is, the point at which the derivative of the function is $0$ or does not exist.
  
  $f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 12 x + 13\right) = 6 x - 12$
  
  Setting this equal to $0$, we find a critical point at $x = 2$
  
  $6 x - 12 = 0 \implies x = \frac{12}{6} = 2$
  
  Now, we can either test to see whether it is an extremum (and what type) by checking some values of $f$ around that point, or by using the second derivative test. Let's use the latter.
  
  $\frac{{d}^{2} x}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left(6 x - 12\right) = 6$
  
  As $f ' ' \left(2\right) = 6 > 0$, the second derivative test tells us that $f \left(x\right)$ has a local minimum at $x = 2$
  
  Thus, using $f ' \left(x\right)$ and $f ' ' \left(x\right)$, we find that $f \left(x\right)$ has a minimum at $x = 2$, matching the result we found using algebra.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer. The question asks about the extrema of the function \( f(x) = 3x^2 - 12x + 13 \) on the interval \((- \infty, \infty)\). The answer provides a step-by-step solution using calculus, including finding the critical points and using the second derivative test to confirm the nature of the extrema.
  
  Label: question-and-answer