2022 ONE(b)

Achievement
Question
Find the xx-value(s) of any stationary points on the graph of the function f(x)=x2+1xf(x)=\dfrac{x^2+1}{x}.

You must use calculus and show any derivatives that you need to find when solving this problem.
Official Answer
f(x)=x2+1xf(x)=\dfrac{x^2+1}{x}

f′(x)=x⋅(2x)−(x2+1)x2f'(x)=\dfrac{x\cdot(2x)-(x^2+1)}{x^2}

x⋅(2x)−(x2+1)x2=0\dfrac{x\cdot(2x)-(x^2+1)}{x^2}=0

2x2−x2−1=02x^2-x^2-1=0

x2=1x^2=1

x=±1x=\pm 1

OR

f(x)=x+x−1f(x)=x+x^{-1}

f′(x)=1−x−2f'(x)=1-x^{-2}

1−1x2=01-\dfrac{1}{x^2}=0

x2−1=0x^2-1=0

x=±1x=\pm 1
Grading Criteria

Achievement (u)

  • Correct solution with correct derivative
  • Must have both solutions: x=±1x=\pm 1

Merit (r)

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Excellence T1

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Excellence T2

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Video Explanation
2022 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 1:38)
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Exam Paper (2022)Assessment Schedule (2022)
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