2024 THREE(c)

Merit
Question
Find the xx-value(s) of any stationary point(s) on the graph of the function f(x)=x25x+4x2+5x+4f(x)=\dfrac{x^2-5x+4}{x^2+5x+4}.

You must use calculus and show any derivatives that you need to find when solving this problem.

You do not need to determine the nature of any stationary point(s) found.
Official Answer
f(x)=10x240(x2+5x+4)2f'(x)=\dfrac{10x^2-40}{(x^2+5x+4)^2}

For stationary points, f(x)=0f'(x)=0

10x240=010x^2-40=0

x=2x=2 or x=2x=-2
Grading Criteria

Achievement (u)

  • Correct expression for f(x)f'(x).

Merit (r)

  • xx-values of both stationary points found, with evidence of a calculus method.

Excellence T1

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Excellence T2

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Video Explanation
2024 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 44:42)
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