Question 16
A section of the graphs of the first and second derivatives of a function are shown.
Sketch a possible graph of the function on the same axes over the domain $0\le x\le 2\pi$. Explain all reasoning used to produce the sketch.
[3 marks]Work back up the ladder
First decide which curve is the gradient of the other. Then the zeros of $f'$ give the turning points of $f$, and the sign of $f''$ says which kind each one is.
Turning points of $f$ sit where $f'=0$. Points of inflection sit where $f''=0$, which is also where $f'$ has its turning points.
Decide which curve is $f'$, then build $f$
The zeros of each curve line up with the turning points of the other, so zeros alone cannot tell you which is which. Test the sign instead. Choose a curve to be $f'$, slide the tangent along it, and compare its gradient with the height of the other curve.
Choose a curve. If it really is $f'$, the other curve must be its gradient at every $x$.
The gradient of the solid curve equals the height of the dashed curve at every $x$. Where the solid curve falls, the dashed curve is below the axis. So the solid curve is $f'$ and the dashed curve is $f''$.
The gradient of the dashed curve and the height of the solid curve have opposite signs. On $0<x<\frac{2\pi}{3}$ the dashed curve rises, but the solid curve is below the axis. So the dashed curve cannot be $f'$. Try the solid curve.
This step opens once you have chosen the curve that really is $f'$. Everything after it depends on that choice.
That is the shape of $f$. The points of inflection sit at $x=\frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}$, where the dashed curve crosses the axis. Integrating $f'(x)=-3\sin\left(\frac{3x}{2}\right)$ gives $f(x)=2\cos\left(\frac{3x}{2}\right)+c$, so any vertical shift works. Try one.
Sort all four turning points correctly and your sketch of $f$ appears on the grid.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The QCAA’s sample response reads an equation off each curve and works back to $f$. Reasoning from the turning points is also accepted.
Both curves are sinusoids with the same period. Read each one off the grid. The solid curve starts at 0 and goes down to $-3$, so it is a negative sine. The dashed curve starts at its lowest point, $-4.5$, so it is a negative cosine. One full cycle takes $\frac{4\pi}{3}$, so $b=\frac{2\pi}{4\pi/3}=\frac{3}{2}$.
Differentiating the solid curve gives the dashed one, because $-3\times\frac{3}{2}=-4.5$. Going the other way fails, since the dashed curve rises on $0<x<\frac{2\pi}{3}$ while the solid curve is negative there. So the smooth curve is the first derivative:
$$\begin{aligned}y'&=-3\sin\left(\tfrac{3}{2}x\right)\\y''&=-4.5\cos\left(\tfrac{3}{2}x\right)\end{aligned}$$
Marker: correctly identifies the smooth curve as the first derivative
Now work back up one rung. The function whose derivative is a negative sine is a cosine with the same period. Integrating divides by $\frac{3}{2}$, and $-3\times\left(-\frac{2}{3}\right)=2$.
$$\begin{aligned}y&=\int -3\sin\left(\tfrac{3}{2}x\right)dx\\&=2\cos\left(\tfrac{3}{2}x\right)+c\end{aligned}$$
The question asks for a possible graph, so take $c=0$: $y=2\cos\left(\frac{3}{2}x\right)$.
Other reasoning is accepted, such as the coordinates of the turning points. Here $y'=0$ at $x=0,\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ 2\pi$. The dashed curve is negative at 0 and $\frac{4\pi}{3}$, so those are maximums, and positive at $\frac{2\pi}{3}$ and $2\pi$, so those are minimums. If you made a slip in step 1, you can still earn this mark by follow-through.
Marker: provides mathematical reasoning for sketch
Sketch it on the given axes over $0\le x\le2\pi$. It starts at a maximum of 2, falls to a minimum of $-2$ at $\frac{2\pi}{3}$, rises to 2 at $\frac{4\pi}{3}$ and ends at a minimum of $-2$ at $2\pi$. It crosses the axis at the points of inflection, $\frac{\pi}{3},\ \pi$ and $\frac{5\pi}{3}$.
The sample sketch uses $c=0$. Your marks for the sketch and the reasoning can follow through from an earlier slip.
Marker: sketches the function
Each phrase of the question gave you something
You are handed two graphs and have to decide which is the derivative of which before anything else can happen. Then you run the derivative ladder backwards, from $f'$ up to $f$, either by reading equations off the grid and integrating or by locating and classifying turning points. Finally you turn that into a sketch and explain it. It is not complex unfamiliar because the question tells you the curves are the first and second derivatives and asks for a sketch on axes you already have.
If I had to guess where you might lose marks, it would be the reasoning mark. A correct sketch with no words beside it leaves a mark on the table. The 2022 subject report advises giving students chances to compare the shapes of graphs of functions with the graphs of their derivatives, and this question is exactly that. Watch the trap too. Each curve’s zeros line up with the other’s turning points, so matching zeros proves nothing. Check the sign of the dashed curve where the solid curve rises or falls, and write that check down.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it