1. Factor the expression inside modulus

We have

So

We only need to study this on the interval $(0,4)$.

2. Determine the sign of $g(x)=(x-1)(x-3)(x+1)$ on $(0,4)$

Since $x+1>0$ for all $x\in(0,4)$, the sign depends on $(x-1)(x-3)$.

• For $0<x<1$: both $(x-1)$ and $(x-3)$ are negative, so their product is positive. Hence $g(x)>0$.
• For $1<x<3$: one factor positive, one negative, so $g(x)<0$.
• For $3<x<4$: both positive, so $g(x)>0$.

Thus,

Now expand:

Therefore,

Also check continuity at the joining points:

And these match the piecewise values, so $f$ is continuous.

3. Find critical points in each interval

For $0<x<1$ and $3<x<4$

so

In $(0,1)$, we have $x>0$ and $x-2<0$, so

Hence $f$ is strictly decreasing on $(0,1)$.

In $(3,4)$, both $x>0$ and $x-2>0$, so

Hence $f$ is strictly increasing on $(3,4)$.

So there is no interior critical point in $(0,1)$ or $(3,4)$.

For $1<x<3$

so

Set $f'(x)=0$:

Thus,

Among these,

so not in $(1,3)$, while

Hence there is exactly one critical point in $(1,3)$.

Now,

At $x=1+\frac{\sqrt{15}}{3}>1$, we get

so this point is a local maximum.

Thus, inside $(1,3)$ we get:

• local maxima: $1$
• local minima: $0$

4. Check non-differentiable points $x=1$ and $x=3$

These points lie inside $(0,4)$, so they may give local extrema.

At $x=1$

From the left, on $(0,1)$, $f'(x)<0$. From the right, for $1<x<3$, evaluate sign of

Near $x=1$,

So the derivative changes from negative to positive at $x=1$. Hence $x=1$ is a local minimum.

At $x=3$

From the left,

From the right, on $(3,4)$,

So the derivative changes from negative to positive at $x=3$. Hence $x=3$ is also a local minimum.

5. Count local minima and maxima

In $(0,4)$:

• Local minima at $x=1$ and $x=3$ $\Rightarrow m=2$
• Local maximum at $x=1+\frac{\sqrt{15}}{3}$ $\Rightarrow M=1$

Therefore,

6. Compare with stored answer

Derived answer: $3$

Stored correct answer: $3$

They agree.