Graphs & Transformations: Infer f(x) from \(f'(x)\) & \(|f(x)|\) — DHS 2025 H2 Math Prelim Paper 1
What this question tests
Question
The graphs of \(y = \mathrm{f}'(x)\) and \(y = |\mathrm{f}(x)|\) are shown below.

- State the nature of all turning point(s) of the graph of \(y = \mathrm{f}(x)\).
- State the range of values of \(x\) where \(\mathrm{f}\) is decreasing.
- Sketch the graph of \(y = \mathrm{f}(x)\), indicating clearly the equations of the asymptote(s), coordinates of the turning point(s) and the intersections with the axes.
- On the copy of the graph of \(y = |\mathrm{f}(x)|\) in the Printed Answer Booklet, sketch and label a line \(y = kx + 3k\), where \(k\) is a constant. Hence state the range of values of \(k\) for which there is no real solution to the equation \(|\mathrm{f}(x)| = kx + 3k\).
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(a) Using the first derivative test from the graph of \(y = \mathrm{f}'(x)\):
| \(x\) | \(-2^-\) | \(-2\) | \(-2^+\) |
|---|---|---|---|
| \(\dfrac{dy}{dx}\) (value of \(\mathrm{f}'(x)\)) | \(+ve\) | \(0\) | \(-ve\) |
| Shape | \(\nearrow\) | \(-\) | \(\searrow\) |
\(\therefore\) Maximum point at \(x = -2\).
| \(x\) | \(1.5^-\) | \(1.5\) | \(1.5^+\) |
|---|---|---|---|
| \(\dfrac{dy}{dx}\) (value of \(\mathrm{f}'(x)\)) | \(-ve\) | \(0\) | \(+ve\) |
| Shape | \(\searrow\) | \(-\) | \(\nearrow\) |
\(\therefore\) Minimum point at \(x = 1.5\).
Maximum point at \(x = -2\); minimum point at \(x = 1.5\).
(b) \(\mathrm{f}\) is decreasing where \(\mathrm{f}'(x) < 0\), i.e. where the graph of \(y = \mathrm{f}'(x)\) is below the \(x\)-axis.
\(-2 \leq x < 0\) or \(0 < x \leq 1.5\)
(c) Sketch of \(y = \mathrm{f}(x)\):

Key features: asymptotes \(x = 0\) and \(y = x + 3\); maximum \((-2, -2)\); minimum \((1.5, -1)\); \(x\)-intercepts at \((1,0)\) and \((2.5,0)\).
(d) The line \(y = kx + 3k = k(x+3)\) always passes through \((-3,0)\). Varying \(k\) pivots the line about \((-3,0)\).

For no intersection with \(y = |\mathrm{f}(x)|\), the line must pass above or below all intersections. From the diagram: