$y = \mathrm{f}'(x)$ 和 $y = |\mathrm{f}(x)|$ 的图像如下图所示。

1. Locate the turning points of $y = \mathrm{f}(x)$ by finding the $x$-intercepts of $y = \mathrm{f}'(x)$.

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

2. Determine the nature of the turning points using the graphical first derivative test.

At $x = -2$, the curve of $y = \mathrm{f}'(x)$ crosses the $x$-axis from positive to negative. This means the gradient of $\mathrm{f}(x)$ changes from positive to negative, so $x = -2$ is a local maximum.
At $x = 1.5$, the curve of $y = \mathrm{f}'(x)$ comes from a local maximum in the first quadrant (positive) and goes below the $x$-axis (negative). Thus, $\mathrm{f}'(x)$ changes from positive to negative, so $x = 1.5$ is also a local maximum.

1. Relate the decreasing property of the function to its derivative.

A function $\mathrm{f}(x)$ is decreasing when its gradient is negative, which means $\mathrm{f}'(x) < 0$.

Graphically, we need to find the regions where the graph of $y = \mathrm{f}'(x)$ lies strictly below the $x$-axis.

2. Identify the relevant intervals from the graph of $y = \mathrm{f}'(x)$.

Looking at the left graph for $y = \mathrm{f}'(x)$:

• For $x < 0$, the curve crosses the $x$-axis at $x = -2$ and goes down to $-\infty$ as it approaches the vertical asymptote $x = 0$. Thus, $\mathrm{f}'(x) < 0$ for $-2 < x < 0$.
• For $x > 0$, the curve crosses the $x$-axis at $x = 1.5$ and remains below the $x$-axis as it approaches the horizontal asymptote $y = 0$. Thus, $\mathrm{f}'(x) < 0$ for $x > 1.5$.

Combining these gives the range of values where $\mathrm{f}$ is decreasing.

1. Identify the fixed point of the line family and sketch the line

Factorise: $y=kx+3k=k(x+3)$. For every value of $k$ this passes through the fixed point where $x+3=0$, i.e. $(-3,0)$. This point is the $x$-intercept of the oblique asymptote $y=-x-3$ and lies below the curve (the left branch has minimum point $(-2,2)$, so $|\mathrm{f}(x)|\ge 2$ there). On the copy of $y=|\mathrm{f}(x)|$, draw a straight line through $(-3,0)$ and label it $y=kx+3k$.

2. Find the gradients for which the line misses the right branch

For $x>0$, $x+3>0$, so the line $y=k(x+3)$ has the same sign as $k$. Since $|\mathrm{f}(x)|\ge0$, a line with $k<0$ lies entirely below the $x$-axis for $x>0$ and so cannot meet the right branch (which touches the axis at $(1,0)$ and $(2.5,0)$). If $k=0$ the line is the $x$-axis, touching the curve at $x=1$ and $x=2.5$ (solutions exist); if $k>0$ the line rises above the touch-points and cuts the right branch. Hence to miss the right branch we need $k<0$.

3. Find the gradient limit set by the oblique asymptote (left branch) and combine

As $x\to-\infty$ the left branch approaches $y=-x-3$ (slope $-1$), which is exactly the line with $k=-1$. For $k\ge -1$ the line stays below the left branch (for $k=-1$ it is parallel to the asymptote, never reached; for $-1<k<0$ it lies under the valley). For $k<-1$ the line eventually rises above the left branch and cuts it. Combining with $k<0$ from Step 2, the line misses both branches for $-1\le k<0,$ giving no real solution to $|\mathrm{f}(x)|=kx+3k$.