$y=f'(x)$ 和 $y=|f(x)|$ 的图像如下所示。

1. Identify the $x$-coordinates of the turning points from the graph of $y = f'(x)$.

Turning points of $y = \mathrm{f}(x)$ occur where $\mathrm{f}'(x) = 0$.
From the given graph of $y = \mathrm{f}'(x)$, the $x$-intercepts are at $x = -2$ and $x = 1.5$.
Thus, the turning points are at $x = -2$ and $x = 1.5$.

2. Determine the nature of each turning point by observing the sign change of $f'(x)$.

At $x = -2$, the graph of $y = \mathrm{f}'(x)$ crosses the $x$-axis from above to below. This means $\mathrm{f}'(x)$ changes from positive to negative, indicating a local maximum.

At $x = 1.5$, the graph of $y = \mathrm{f}'(x)$ also falls and crosses the $x$-axis from above to below. This means $\mathrm{f}'(x)$ changes from positive to negative, indicating a local maximum as well.

1. State the condition for f to be decreasing in terms of f'.

$f$ is decreasing where $f'(x) < 0$, i.e. where the graph of $y=f'(x)$ lies below the $x$-axis.

2. Read the interval where f'(x) < 0 from the graph.

From the graph of $y=f'(x)$, the curve is below the $x$-axis between its zeros at $x=-2$ and $x=1.5$ (it is positive for $x<-2$). Hence $f'(x)<0$ for $-2 < x < 1.5$.

1. Identify the asymptotes of f(x)

Since $y=|f(x)|$ has a vertical spike at $x=1.5$, $f(x)$ has a vertical asymptote $x=1.5$.

As $x\to-\infty$, $f'(x)\to1$, so $f$ has gradient $1$ far out; the reference lines $y=x+3$ and its reflection $y=-x-3$ on the $|f(x)|$ graph (meeting the axis at $(-3,0)$) show the oblique asymptote $y=x+3$.

2. Recover the sign of f and locate the turning point

$f'(x)=0$ at $x=-2$ with $f'>0$ for $x<-2$ and $f'<0$ just after, so $x=-2$ is a maximum. From the $|f(x)|$ graph the point $(-2,2)$ lies on a smooth peak, and $f$ stays positive on $-3<x<1$, so the maximum is at $(-2,2)$.

The parts of $|f(x)|$ that touch the axis with a corner ($x=-3,\,1,\,2.5$) are where $f=0$. Reflect back below the axis the sections where $f$ was negative (for $x<-3$, for $1<x<1.5$, and for $1.5<x<2.5$) to obtain the true graph of $f$.

3. Produce the labelled sketch with all required features

Draw both asymptotes $x=1.5$ (dashed vertical) and $y=x+3$ (dashed oblique). Plot the maximum $(-2,2)$ and the intercepts $(-3,0)$, $(1,0)$, $(2.5,0)$.

Left branch: comes up along $y=x+3$ from the lower left, crosses $(-3,0)$, rises to the maximum $(-2,2)$, then falls, crossing $(1,0)$ and dropping to $-\infty$ as $x\to1.5^-$.

Right branch: rises from $-\infty$ as $x\to1.5^+$, crosses $(2.5,0)$ and approaches the oblique asymptote $y=x+3$ from below.