2024 TWO(d)

Merit
Question
Consider the function f(x)=lnxxf(x)=\dfrac{\ln x}{x}, x>0x>0.

Find the coordinates of the point of inflection on the graph of the function.

*You can assume that your point found is actually a point of inflection.*

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
f(x)=1lnxx2f'(x)=\dfrac{1-\ln x}{x^2}
f(x)=3+2lnxx3f''(x)=\dfrac{-3+2\ln x}{x^3}
For inflection f(x)=0f''(x)=0
giving 3+2lnx=0-3+2\ln x=0
lnx=32\ln x=\dfrac{3}{2}
x=e1.5=4.4817x=e^{1.5}=4.4817
y=1.5e1.5=0.3347y=1.5e^{-1.5}=0.3347
i.e. (4.4817,0.3347)(4.4817,0.3347)
Grading Criteria

Achievement (u)

  • Correct expression for dydx\dfrac{dy}{dx}.

Merit (r)

  • Correct co-ordinates for point of inflection, with evidence of both dydx\dfrac{dy}{dx} and d2ydx2\dfrac{d^2y}{dx^2}.

Excellence T1

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Excellence T2

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Video Explanation
2024 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 27:36)
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