Question 16
Consider the following graph of $f(x)$.
Identify the graph of the second derivative $f''(x)$ from the graphs in Diagram 1, Diagram 2 and Diagram 3.
Justify your decisions using mathematical reasoning.
[4 marks]Take the gradient twice
Watch a tangent slide along the graph from the question. Its gradient draws the first derivative, and doing it again draws the second.
The second derivative crosses the $x$-axis where $f(x)$ changes concavity. It is positive at the minimums of $f(x)$ and negative at the maximums.
Test each diagram against the concavity of $f(x)$
Concave up means $f''(x)>0$. Concave down means $f''(x)<0$. At a point of inflection, $f''(x)=0$. Pick a diagram, then move $x$ along the graph. One point where the signs disagree is enough to rule a diagram out.
Diagram 2 sits near zero at the turning points of $f(x)$ and is above the axis where $f(x)$ is increasing. That is close to how $f'(x)$ behaves, so Diagram 2 looks like a first derivative.
Diagram 3 is zero at the turning points of $f(x)$ as well, but it is below the axis where $f(x)$ is increasing. It looks like $f'(x)$ flipped upside down, so it matches neither derivative.
Diagrams 2 and 3 are both ruled out. Diagram 1 agrees with the concavity of $f(x)$ at every point. It crosses the $x$-axis at the points of inflection, $x\approx 0.75,\ 2.25,\ 3.75,\ 5.25$, so Diagram 1 is $f''(x)$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA’s sample response, which reads $f'(x)$ from the graph first and then reads $f''(x)$ from $f'(x)$.
Choose a route before you look at the diagrams. $f''(x)$ is the gradient of $f'(x)$, so go one derivative at a time. Where $f(x)$ increases and decreases tells you the sign of $f'(x)$. Then where $f'(x)$ increases and decreases tells you the sign of $f''(x)$.
Start by reading the turning points off the graph. $f(x)$ has minimums at $x=0$, $3$ and $6$ and maximums at $x=1.5$ and $4.5$. So $f(x)$ increases and then decreases between $x=0$ and $x=3$, and does the same again between $x=3$ and $x=6$. In between, it changes concavity.
Other methods are accepted. You could use calculus on a model of $f(x)$, sketch $f'(x)$, find the points of inflection, or use the nature and location of the turning points.
Marker: correctly identifies an appropriate method to determine $f''(x)$ graph
Turn the shape of $f(x)$ into the sign of $f'(x)$. Increasing means a positive gradient, decreasing means a negative one, and a turning point means zero. $f(x)$ increases for $0<x<1.5$ and decreases for $1.5<x<3$, so:
$$\begin{aligned}0<x<1.5&:\ f'(x)>0\\1.5<x<3&:\ f'(x)<0\\x=0,\ 1.5,\ 3&:\ f'(x)=0\end{aligned}$$
The same pattern repeats from $x=3$ to $x=6$. So the sketch of $f'(x)$ is above and below the $x$-axis on these intervals.
Other reasoning that fits your method is accepted.
Marker: correctly determines a relevant feature of the graph of $f(x)$
Now read your sketch of $f'(x)$ the same way. It has a maximum where $f(x)$ is rising most steeply and a minimum where $f(x)$ is falling most steeply. Each one sits halfway between two turning points of $f(x)$.
$$\begin{aligned}f'(x)\text{ max at }x&\approx 0.75\\f'(x)\text{ min at }x&\approx 2.25\end{aligned}$$
At a maximum or minimum of $f'(x)$, its gradient is zero. So the sketch of $f''(x)$ crosses the $x$-axis at these points, and again at $x\approx 3.75$ and $5.25$. These are the points of inflection of $f(x)$.
You are reading from a graph, so approximate values are fine. Readings such as $x\approx 0.8$ and $x\approx 2.4$ still describe the same feature. Other reasoning that fits your method is accepted.
Marker: correctly determines a relevant feature for $f''(x)$
Match the features to the diagrams. Diagram 1 crosses the $x$-axis at about $x=0.75,\ 2.25,\ 3.75$ and $5.25$. It is positive at the minimums of $f(x)$ and negative at the maximums, which fits concave up and concave down.
Diagrams 2 and 3 cross the $x$-axis near $x=0,\ 1.5,\ 3,\ 4.5$ and $6$, the turning points of $f(x)$. That is where $f'(x)$ is zero. Diagram 2 looks like your sketch of $f'(x)$ and Diagram 3 looks like that sketch upside down. Neither one can be $f''(x)$. At $x=1.5$, for example, $f(x)$ is concave down so $f''(x)$ must be negative, but both diagrams are close to zero there.
Marker: correctly identifies Diagram 1
Each phrase of the question gave you something
You are given a graph and asked for its second derivative, which is the graph reading you practise in this topic. The link between $f(x)$, $f'(x)$ and $f''(x)$ does all the work, and the turning points are easy to read off the grid. Nothing needs to be calculated.
If I had to guess where you might lose marks, it would be the justification. Choosing Diagram 1 is worth one mark of the four. The QCAA’s subject report says effective responses selected the diagram and gave a justification for the selection. Name the features you used, such as the turning points of $f(x)$ and the points of inflection where $f''(x)=0$. Also watch for Diagram 2. It is the closest match to $f'(x)$, so it is easy to pick if you only take one derivative.
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 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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