Animated Solution for Mathematics - Differentiation: Let $f(x)=|(x-1)(x^2-2x-3)|+x-3,x\in \mathbb{R}$. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ______

Unlocking the Modulus

Welcome, fellow explorer of the mathematical landscape. Today, we are not just solving a problem; we are dissecting a function.

We are looking at $f(x)=|(x-1)(x^2-2x-3)|+x-3$. At first glance, this might look like a chaotic mess of absolute values and polynomials.

Imagine this function as a piece of wire that you are bending. The modulus creates sharp corners, and the polynomial terms create smooth curves. Our goal is to map this terrain and find every peak and valley.

Before we touch any calculus, we must understand the structure. The expression inside the modulus, $(x-1)(x^2-2x-3)$, is the heart of the problem.

Let's factorize it. We know that $x^2-2x-3$ is simply $(x-3)(x+1)$.

So, our function becomes:

Notice the symmetry? We can group $(x-1)(x+1)$ to get $(x^2-1)$. Thus, our function is:

The 'hinges'—the points where the expression inside the modulus flips sign—are at $x=-1,1,3$. Since our domain is restricted to $(0,4)$, we only care about the hinges at $x=1$ and $x=3$. These are the points where our graph will likely change its behavior drastically.

To analyze this, we must break the function into three distinct regions: $(0,1)$, $(1,3)$, and $(3,4)$.

In the first interval, $x\in (0,1)$, both $(x^2-1)$ and $(x-3)$ are negative. A negative times a negative is positive, so the modulus opens with a positive sign.

We get:

In the middle interval, $x\in (1,3)$, $(x^2-1)$ is positive, but $(x-3)$ is negative. Their product is negative, so the modulus must open with a negative sign.

We get:

Finally, in the interval $x\in (3,4)$, both terms are positive, so the modulus opens positively again, returning us to:

Now, we differentiate. For the outer intervals, $f^{\prime }(x)=3x^2-6x$. For the middle interval, $f^{\prime }(x)=-3x^2+6x+2$.

Let's test the hinges. At $x=1$, the Left-Hand Derivative (LHD) is $3(1{)}^2-6(1)=-3$. The Right-Hand Derivative (RHD) is $-3(1{)}^2+6(1)+2=5$.

The slope jumps from negative to positive. This is a classic local minimum! The same logic applies at $x=3$. The LHD is $-7$ and the RHD is $9$. Again, the slope jumps from negative to positive. Another local minimum!

Are we done? Not quite. We must check the middle interval $(1,3)$ for any smooth peaks.

Setting $f^{\prime }(x)=-3x^2+6x+2=0$, we use the quadratic formula to find:

Only $x=1+\sqrt{\frac{5}{3}}\approx 2.29$ lies within our interval. A quick check with the second derivative, $f^{\prime \prime }(x)=-6x+6$, shows that at $x\approx 2.29$, $f^{\prime \prime }(x)<0$. This confirms a local maximum.

We have found two local minima and one local maximum. The sum $m+M=2+1=3$. We have successfully navigated the terrain of this function.