Question 3(d) Merit

A log chain rule, three readings off a drawn graph, a parametric normal on a Ferris wheel, stationary points of a reciprocal pair, and a catenary to verify by substitution.

Find the co-ordinates of any stationary points on the graph of f(x) = 1/x − 2/x³, identifying their nature.

The 2023 paper, page 12 ↗ Marking schedule, page 6 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1. 1 Write f with negative powers, and differentiate.

   f(x) = x⁻¹ − 2x⁻³
   f'(x) = −x⁻² + 6x⁻⁴ = (6 − x²)/x⁴

2. 2 Set it to zero.

   6 − x² = 0, so x = ±√6 ≈ ±2.449. (x = 0 is not in the domain.)

3. 3 Find the y-coordinates.

   f(√6) = 1/√6 − 2/6√6 = 2/3√6 = √6/9 ≈ 0.272
   f is odd, so f(−√6) = −√6/9.

4. 4 Use the second derivative to name each one.

   f''(x) = 2x⁻³ − 24x⁻⁵
   f''(√6) ≈ −0.136 < 0: a maximum.
   f''(−√6) ≈ 0.136 > 0: a minimum.

What the marker wanted

• Achieved Both values of x from a correct derivative, with x = 0 left out.
• Merit Both stationary points found, and each one named, maximum or minimum, by a first- or second-derivative test.

The schedule marks this part twice, once per method. One version asks for the coordinates; the other says the y-values are not needed for Merit. The question asks for co-ordinates, so give them.

In our words, from NZQA’s 2023 marking schedule.

How did it go?

Got it moves it further out. Review brings it back tomorrow.