AS 91578 · 2023 paper · gradient of normal, turning points, a proof
Question 3(b)(i) Achieved
Practises Reading a graph: differentiability, f′ = 0, and limits
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.
The graph shows the function y = f(x).
Find the value(s) of x where f(x) is continuous but not differentiable.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 Continuous means unbroken. Where does the graph break?
x = −2: the curve and the flat line both stop at an open circle, so f(−2) is not defined.
x = 2: the flat line and the U-shaped curve meet at an open circle too.
x = 6: the curve ends at a filled dot at 2, and the next piece starts from an open circle at 5. A jump. -
2 Where is the graph unbroken, but sharp?
At x = 8, the bottom of the V. The graph is joined there, but the gradient jumps from negative to positive.
What the marker wanted
- Achieved Two of the three answers to (i), (ii) and (iii) correct.
The schedule marks (i), (ii) and (iii) together as one Achieved item: any two of the three correct earns it.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.