NCEA Calculus 2023 THREE(d - Solution & Explanation

Question

Find the co-ordinates of any stationary points on the graph of the function

f(x)=\frac{1}{x}-\frac{2}{x^3}

,
identifying their nature.

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

Official Answer

f(x)=x^{-1}-2x^{-3}

f'(x)=-x^{-2}+6x^{-4}

=-\frac{1}{x^2}+\frac{6}{x^4}

f'(x)=0\Rightarrow-\frac{1}{x^2}+\frac{6}{x^4}=0

(x^4-6x^2)=0

x^2(x^2-6)=0

x=0

not possible

x^2-6=0

x=\pm\sqrt{6}

or

x=\pm 2.45

Second derivative test :

f''(x)=2x^{-3}-24x^{-5}

=\frac{2}{x^3}-\frac{24}{x^5}

f''(\sqrt{6})=-0.136=-\frac{\sqrt{6}}{18}

Since

f''(\sqrt{6})<0

,

x=\sqrt{6}

is a local maximum.

f''(-\sqrt{6})=0.136=\frac{\sqrt{6}}{18}

Since

f''(-\sqrt{6})>0

,

x=-\sqrt{6}

is a local minimum.
Maximum turning point when

x=\sqrt{6}

.
Minimum turning point when

x=-\sqrt{6}

.

Grading Criteria

$x$ -coordinates of both stationary points found.
AND
The nature of the two turning points found with a correct first or second derivative test.

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Video Explanation

2023 NCEA L3 Calculus Exam Walkthrough by infinityplusone(starts at 58:35)

2023 THREE(d)