Question 3
The derivative of the function $f(x)$ is given by $f'(x)=\sin\left(x^3\right)$ for the domain $-1.8\lt x\lt1.8$.
The number of points of inflection that the graph of $f(x)$ has on this interval is
Not quite. $f''(x)=3x^2\cos\left(x^3\right)$ changes sign more than once on this domain. Graph it over $-1.8\lt x\lt1.8$ and count every place it crosses the axis. In 2022, 23.1% of Queensland students chose (A).
Not quite. 3 is the number of stationary points: $f'(x)=\sin\left(x^3\right)=0$ at $x=0$ and $x=\pm\sqrt[3]{\pi}\approx\pm1.465$. Points of inflection come from the second derivative changing sign. In 2022, 27.8% of Queensland students chose (B).
That’s it. Step through the hints below to see why. QCAA accepted both (C) and (D) for this question; the note under the solution explains why (C) is the mathematically correct count. In 2022, 21.1% of Queensland students chose (C).
QCAA accepted this when marking, but it isn’t the mathematically correct count. 5 counts every solution of $f''(x)=0$, including $x=0$. But $f''(x)=3x^2\cos\left(x^3\right)$ is positive on both sides of $x=0$, so the concavity doesn’t change there and it isn’t a point of inflection. In 2022, 27.2% of Queensland students chose (D).
Here $f''(0)=0$, but $f''(x)$ is positive on both sides: the blue curve touches the axis and turns back. The concavity doesn’t change, so $x=0$ isn’t a point of inflection. On the top graph, $f'(x)$ flattens out but keeps rising.
Here $f''(x)=0$ and it changes sign, so the graph of $f$ goes from to . On the top graph, $f'(x)$ has a turning point. This is a point of inflection.
Here $f''(x)\gt0$, so the graph of $f$ is concave up, and $f'(x)$ is increasing. Slide across the whole domain and count where the blue curve crosses the axis.
Here $f''(x)\lt0$, so the graph of $f$ is concave down, and $f'(x)$ is decreasing. Slide across the whole domain and count where the blue curve crosses the axis.
Points of inflection are where the concavity changes, and concavity comes from the second derivative. You’re given $f'(x)$, so differentiate once with the chain rule.
$$f''(x)=3x^2\cos\left(x^3\right)$$
Set it equal to zero. Either $3x^2=0$ or $\cos\left(x^3\right)=0$.
$x^3=\pm\dfrac{\pi}{2}\ \text{or}\ {\pm}\dfrac{3\pi}{2}$
The next one, $\tfrac{5\pi}{2}\approx7.85$, is too big, because $1.8^3\approx5.83$. That gives five candidates:
$$x=0,\ \pm1.162,\ \pm1.677$$
A zero of $f''(x)$ is only a candidate. It’s a point of inflection only if $f''(x)$ changes sign there.
At $\pm1.162$ and $\pm1.677$ the cosine changes sign, so $f''(x)$ does too. At $x=0$, $3x^2$ is never negative and $\cos\left(x^3\right)$ is close to 1, so $f''(x)$ is positive on both sides. No sign change.
In the figure, the blue curve crosses the axis four times and only touches it at $x=0$.
$\displaystyle f''(x)=3x^2\cos\left(x^3\right)=0$$\displaystyle \Rightarrow\ x=0,\ \pm1.162,\ \pm1.677$
$f''(x)$ changes sign at $\pm1.162$ and $\pm1.677$, but not at $x=0$, where it is positive on both sides. So there are 4 points of inflection.
A note on the key. QCAA’s published marking guide lists (D) for this question, and both (C) and (D) were accepted when the exam was marked. The count of 5 includes $x=0$, where $f''(0)=0$ but the concavity doesn’t change. A point of inflection needs a change of concavity, so this page gives (C).
How Queensland students answered in 2022
The QCAA also gave the mark for (D), so 48.3% of students got it. From the QCAA Mathematical Methods subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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