The graph of $y = \text{f}(x)$ has a maximum turning point at $(2,3)$ and passes through the origin. The lines $x = -1$ and $y = 2$ are asymptotes to the graph, as shown in the diagram below.

1. Relate the increasing behaviour of $\frac{1}{\mathrm{f}(x)}$ to that of $\mathrm{f}(x)$

Differentiating, $\dfrac{d}{dx}\!\left(\dfrac{1}{\mathrm{f}(x)}\right) = -\dfrac{\mathrm{f}'(x)}{[\mathrm{f}(x)]^2}$.

Since $[\mathrm{f}(x)]^2 > 0$, the reciprocal graph is increasing precisely where $\mathrm{f}'(x) < 0$, i.e. where $y=\mathrm{f}(x)$ is decreasing.

2. Identify where $\mathrm{f}$ is decreasing from the given graph

From the diagram:

• For $x<-1$, the curve decreases as $x$ increases (and stays above $y=2$, so $\mathrm{f}\neq 0$).
• For $-1<x<2$, the curve increases up to the maximum at $(2,3)$.
• For $x>2$, the curve decreases towards $y=2$ (staying above $2$, so $\mathrm{f}\neq 0$).

Thus $\mathrm{f}$ is decreasing for $x<-1$ and for $x>2$.

3. State the required range

Therefore $y=\dfrac{1}{\mathrm{f}(x)}$ is increasing for
$x<-1 \quad \text{or} \quad x>2.$

1. Determine the sign of f'(x) on each interval from the shape of f

From the graph of $y=\text{f}(x)$:

• For $x<-1$, $f$ is decreasing $\Rightarrow f'(x)<0$.
• For $-1<x<2$, $f$ is increasing (from $-\infty$ up to the maximum) $\Rightarrow f'(x)>0$.
• At the maximum turning point $x=2$, $f'(x)=0$.
• For $x>2$, $f$ is decreasing (towards $y=2$) $\Rightarrow f'(x)<0$.

2. Find the intercepts and asymptotes of y = f'(x)

$x$-intercept: $f'(2)=0$, so the graph cuts the $x$-axis at $(2,0)$.

$y$-intercept: at the origin $f$ is increasing, so $f'(0)>0$ — a positive $y$-intercept.

Asymptotes: $f$ has a vertical asymptote at $x=-1$, so $f'$ also has the vertical asymptote $x=-1$. Since $f\to 2$ (a constant) as $x\to\pm\infty$, the gradient $f'(x)\to 0$, giving the horizontal asymptote $y=0$.

3. Translate limiting slopes into the two branches and sketch

Limiting gradients:

• $x\to-\infty$: $f$ flattens onto $y=2$ from above while decreasing $\Rightarrow f'(x)\to 0^-$.
• $x\to-1^-$: $f\to+\infty$ steeply (decreasing) $\Rightarrow f'(x)\to-\infty$.
• $x\to-1^+$: $f\to-\infty$ but rising steeply $\Rightarrow f'(x)\to+\infty$.
• $x\to\infty$: $f$ approaches $y=2$ from above (decreasing) $\Rightarrow f'(x)\to 0^-$.

Left branch ($x<-1$): lies below the $x$-axis, rising from $-\infty$ near $x=-1$ to $0^-$ as $x\to-\infty$.
Right branch ($x>-1$): falls from $+\infty$ near $x=-1$, through a positive $y$-intercept, crossing $(2,0)$, then below the axis approaching $0^-$ as $x\to\infty$.