NCEA Calculus 2022 THREE(d - Solution & Explanation

Question

Find the

x

-value(s) of any stationary point(s) on the graph of the function

y = 9x - 2 + \dfrac{3}{3x - 1}

and
determine their nature.

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

Official Answer

y=9x-2+\dfrac{3}{3x-1}

\dfrac{dy}{dx}=9-3(3x-1)^{-2}\times 3

=9-\dfrac{9}{(3x-1)^2}

Stationary point

\dfrac{dy}{dx}=0

9-\dfrac{9}{(3x-1)^2}=0

9=\dfrac{9}{(3x-1)^2}

(3x-1)^2=1

3x-1=\pm 1

x=\dfrac{1\pm 1}{3}

x=0

or

x=\dfrac{2}{3}

\dfrac{d^2y}{dx^2}=\dfrac{54}{(3x-1)^3}

x=0\quad \dfrac{d^2y}{dx^2}=\dfrac{54}{(-1)^3}<0

Local max at

x=0

x=\dfrac{2}{3}\quad \dfrac{d^2y}{dx^2}=\dfrac{54}{(1)^3}>0

Local min at

x=\dfrac{2}{3}

Grading Criteria

Correct solution with correct derivative.
The nature of each turning point stated but not determined using a calculus method.

T1 Correct solution with correct derivative.
The nature of each turning point determined with a first or second derivative test.

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Video Explanation

2022 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 62:31)

2022 THREE(d)