The diagrams below show the graphs of $y = |\mathrm{f}(x)|$ and $y = \mathrm{f}'(x)$. On separate diagrams, sketch the graphs of: showing clearly, in each case, the intersection(s) with the axes, the coordinates of the turning point(s) and the equation(s) of the asymptotes.

1. Identify the geometrical transformations applied to the original graph $y = |\mathrm{f}(x)|$.

To obtain the graph of $y = |\mathrm{f}(2x)| + 1$ from $y = |\mathrm{f}(x)|$, we apply the following sequence of transformations:

1. A horizontal scaling parallel to the $x$-axis by a scale factor of $\frac{1}{2}$.
2. A vertical translation of $1$ unit in the positive $y$-direction.

In terms of coordinates, any point $(x, y)$ on the original graph is mapped to $\left(\frac{x}{2}, y + 1\right)$.

2. Determine the new equations for the asymptotes.

Apply the transformations to the original asymptotes:

• Horizontal asymptote: $y = 2$ is scaled horizontally (which has no effect) and then translated upwards by $1$. The new horizontal asymptote is $y = 2 + 1 \implies y = 3$.
• Vertical asymptote: $x = 1$ is scaled horizontally by a factor of $\frac{1}{2}$ and translated vertically (which has no effect). The new vertical asymptote is $x = 1 \times \frac{1}{2} \implies x = \frac{1}{2}$.

3. Calculate the transformed coordinates of the intersections and turning points.

Apply the mapping $(x, y) \to \left(\frac{x}{2}, y + 1\right)$ to the key points of $y = |\mathrm{f}(x)|$:

• $y$-intercept: $(0, 1) \to (0, 2)$.
• First minimum: $\left(\frac{1}{2}, 0\right) \to \left(\frac{1}{4}, 1\right)$.
• Second minimum: $(2, 0) \to (1, 1)$.
• Local maximum: $(5, 3) \to \left(\frac{5}{2}, 4\right)$.

Since the minimum $y$-value of the new graph is $1$, the graph lies entirely above the $x$-axis, meaning there are no $x$-intercepts.

1. Find the intersection points with the coordinate axes.

To find the $y$-intercept, substitute $x = 0$:
$y = \frac{1}{\mathrm{f}'(0)} = \frac{1}{-3} = -\frac{1}{3}$
So, the $y$-intercept is $(0, -\frac{1}{3})$.

To find the $x$-intercept, set $y = 0$:
$\frac{1}{\mathrm{f}'(x)} = 0 \implies \mathrm{f}'(x) \to \pm\infty$
From the given graph of $\mathrm{f}'(x)$, this occurs at the vertical asymptote $x = 1$. Thus, the curve intersects the $x$-axis at $(1, 0)$.

2. Identify the asymptotes.

Vertical asymptotes occur where the denominator is zero, i.e., $\mathrm{f}'(x) = 0$. From the given graph, the $x$-intercept is at $x = 5$. Hence, there is a vertical asymptote at $x = 5$.

As $x \to \pm\infty$, the graph shows $\mathrm{f}'(x) \to 0$. As the denominator approaches $0$, the value of $y = \frac{1}{\mathrm{f}'(x)} \to \pm\infty$. Thus, there are no horizontal asymptotes.

3. Find the coordinates and nature of the turning points.

Turning points of $y = \frac{1}{\mathrm{f}'(x)}$ correspond to turning points of $\mathrm{f}'(x)$.
The given graph has a local minimum at $(8, -3)$.
The corresponding point on the reciprocal graph is $\left(8, \frac{1}{-3}\right) = \left(8, -\frac{1}{3}\right)$.
Since $\mathrm{f}'(x)$ is negative and has a local minimum at this point, taking its reciprocal produces a local maximum. Thus, there is a local maximum at $\left(8, -\frac{1}{3}\right)$.

4. Deduce the shape of the graph and sketch.

Synthesising the features for the sketch:

• For $x < 1$: $\mathrm{f}'(x)$ is negative and approaches $-\infty$ from $0$. Thus $y$ increases from $-\infty$ to $0$, crossing the $y$-axis at $(0, -1/3)$.
• For $1 < x < 5$: $\mathrm{f}'(x)$ is positive, decreasing from $\infty$ to $0$. Thus $y$ is positive, increasing from $(1, 0)$ up towards $\infty$ at $x=5^-$.
• For $x > 5$: $\mathrm{f}'(x)$ is negative, goes down to $-3$ and back up to $0$. Thus $y$ is negative, comes up from $-\infty$ at $x=5^+$, reaches a local maximum at $(8, -1/3)$, and curves back down towards $-\infty$ as $x \to \infty$.

1. Determine the shape and sign of $f(x)$ for $x < 1$.

From the graph of $y = \mathrm{f}'(x)$, we observe that $\mathrm{f}'(x) < 0$ for $x < 1$. Thus, $\mathrm{f}(x)$ is strictly decreasing on $x < 1$.
The graph of $y = |\mathrm{f}(x)|$ touches the $x$-axis at $x = 0.5$, which implies $\mathrm{f}(0.5) = 0$.
Since $\mathrm{f}(x)$ is strictly decreasing, we deduce that $\mathrm{f}(x) > 0$ for $x < 0.5$ and $\mathrm{f}(x) < 0$ for $0.5 < x < 1$.
Therefore, $\mathrm{f}(x) = |\mathrm{f}(x)|$ for $x < 0.5$, and $\mathrm{f}(x) = -|\mathrm{f}(x)|$ for $0.5 < x < 1$. The curve reflects below the $x$-axis for $0.5 < x < 1$, approaching $-\infty$ as $x \to 1^-$.

2. Determine the shape and sign of $f(x)$ for $x > 1$.

From the graph of $y = \mathrm{f}'(x)$, $\mathrm{f}'(x) > 0$ for $1 < x < 5$ and $\mathrm{f}'(x) < 0$ for $x > 5$. Thus, $\mathrm{f}(x)$ is strictly increasing on $(1, 5)$ and strictly decreasing on $x > 5$.
The graph of $y = |\mathrm{f}(x)|$ touches the $x$-axis at $x = 2$, implying $\mathrm{f}(2) = 0$.
Since $\mathrm{f}(x)$ is increasing on $(1, 5)$, we deduce that $\mathrm{f}(x) < 0$ for $1 < x < 2$ and $\mathrm{f}(x) > 0$ for $x > 2$.
Therefore, $\mathrm{f}(x) = -|\mathrm{f}(x)|$ for $1 < x < 2$ (approaching $-\infty$ as $x \to 1^+$), and $\mathrm{f}(x) = |\mathrm{f}(x)|$ for $x > 2$.
At $x = 5$, $\mathrm{f}(x)$ reaches a local maximum.

3. Identify all key features to be labeled on the sketch.

To sketch $y = \mathrm{f}(x)$, draw the curve identical to $y = |\mathrm{f}(x)|$ for $x < 0.5$ and $x > 2$, and reflect the sections $0.5 < x < 1$ and $1 < x < 2$ across the $x$-axis.
The required features to be clearly labeled on the sketch are:

• $y$-intercept: $(0, 1)$
• $x$-intercepts: $(0.5, 0)$ and $(2, 0)$ (the curve smoothly crosses the $x$-axis at both points without sharp turns)
• Turning point: Local maximum at $(5, 3)$
• Asymptotes: $x = 1$ (vertical) and $y = 2$ (horizontal)