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wonk

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Dec 6th, 2019
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Java 0.64 KB | None | 0 0
  1.         steps1.add(new StepsHelper("Step 5", "if \\(0<t<3)\\) then \\(f'(x) = 0\\) at the point \\(x = \\sqrt{t/3}\\), which is on the interval \\(0,1)\\)."
  2.                 + "At this point, <br/>"
  3.                 + "\\(f(\\sqrt{t/3}) = \\frac{t}{3}*\\sqrt{t/3}-t*\\sqrt{t/3}+t\\)<br/>"
  4.                 + "\\(= -\\frac{2t}{3}\\sqrt{t/3}+t\\)<br/>"
  5.                 + "\\(= t(-\\frac{2}{3}\\sqrt{t/3}+1)\\)<br/>"
  6.                 + "Since \\(\\sqrt{t/3} < 1 \\) and \\(t > 0\\), this quantity is positive. Since \\(f(t)\\) is decreasing to on one side of the critical point"
  7.                 + "and increasing on the other,<br/> there can be no solutions on the interval \\((0,\\sqrt{t/3})\\) or on the interval \\(\\sqrt{t/3},1)\\)"));
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