Question 5
The graph of $f''(x)$ is shown.
Which of the following could be the graph of $f'(x)$?
Not quite. (A) is made of two straight lines, so its gradient is a constant on each side of $x=0$. The second derivative would then be flat pieces, not a parabola. Its signs don’t fit either: between $x=-1$ and $x=0$, (A) is increasing, but $f''(x)\lt0$ there. In 2021, 5.2% of Queensland students chose (A).
Not quite. (B) has the same sign pattern as the given graph: negative between $-1$ and 1, positive outside. That describes $f''(x)$ itself, not its antiderivative. Its gradient is a constant on each side of $x=0$, so differentiating (B) gives flat pieces, not a parabola. In 2021, 9.0% of Queensland students chose (B).
Close. (C) turns at $x=-1$ and $x=1$, where $f''(x)=0$, but both turns are the wrong way round. For $x\lt-1$, $f''(x)\gt0$, so $f'$ must be increasing, and (C) is decreasing there. Its gradient has the opposite sign to the given graph everywhere. In 2021, 20.9% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2021, 64.6% of Queensland students chose (D).
Here $f''(x)$ is , so $f'$ must be . is here too, so it passes at this $x$.
Here $f''(x)$ is , so $f'$ must be . But is here, so it can’t be the graph of $f'(x)$.
Here $f''(x)=0$ and it changes from , so $f'$ has a local . (D) has a local right here.
(C) does turn here, but it’s the wrong kind of turn. $f''(x)$ changes from , so $f'$ needs a local . (C) has a local .
Here $f''(x)=0$ and it changes sign, so $f'$ needs a turning point. is a straight line here, so it doesn’t turn.
has a sharp corner at $x=0$, so it has no gradient there. But the given graph has a value at $x=0$, so $f'$ has a gradient there and can’t have a corner.
$f''(x)$ is the gradient of $f'(x)$. So read the given graph as a gradient graph:
$f''(x)\gt0:\ f'\ \text{increasing}$$f''(x)\lt0:\ f'\ \text{decreasing}$
The given graph is positive for $x\lt-1$, negative between $-1$ and 1, and positive for $x\gt1$. So $f'$ increases, then decreases, then increases.
That means a local maximum of $f'$ at $x=-1$ and a local minimum at $x=1$.
The given graph is a parabola, so $f'$ is a cubic, one degree higher. (A) and (B) are made of straight lines, so they’re out.
(C) and (D) both turn at $x=-1$ and $x=1$. Only (D) has the maximum on the left and the minimum on the right. Its height doesn’t matter: adding a constant to $f'$ doesn’t change $f''$.
$f''(x)$ is the gradient of $f'(x)$. From the graph, $f''(x)\gt0$ for $x\lt-1$, $f''(x)\lt0$ for $-1\lt x\lt1$, and $f''(x)\gt0$ for $x\gt1$.
So $f'$ increases, has a local maximum at $x=-1$, decreases, has a local minimum at $x=1$, then increases. Since $f''$ is a parabola, $f'$ is a cubic. Only (D) has that shape.
How Queensland students answered in 2021
From the QCAA Mathematical Methods subject report for 2021. 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 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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