NCEA Calculus 2020 THREE(d - Solution & Explanation

Question

The graph of the function

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

,

x \ne 3

, has two stationary points.

Find the

x

-coordinates of the stationary points, and determine whether they are local maxima
or local minima.

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

Official Answer

y=(x-3)^{-1}+x

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

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

\dfrac{dy}{dx}=0\Rightarrow x-3=\pm 1

x=2\text{ or }4

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

x=2\Rightarrow \dfrac{d^2y}{dx^2}<0\ \text{Local max at }x=2

x=4\Rightarrow \dfrac{d^2y}{dx^2}>0\ \text{Local min at }x=4

Grading Criteria

Achievement (u)

Correct expression for $\dfrac{dy}{dx}$ .

Merit (r)

Correct expressions for $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$
OR
Correct expression for $\dfrac{dy}{dx}$ plus $x$ -coordinates of TPs found and nature stated without correct use of first or second derivative test.

Excellence T1

Correct solution with correct derivatives. With use of the first derivative test or second derivative test to justify the nature of the turning points.

Excellence T2

-

Video Explanation

NCEA Level 3 Calculus Differentiation 2020 NZQA Exam - Worked Answers by infinityplusone(starts at 48:15)

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2020 THREE(d)

Achievement (u)

Merit (r)

Excellence T1

Excellence T2