Animated Solution for Mathematics - Differentiation: Let $f(x)=\{\begin{array}{ll}-x^3+\frac{b^3-b^2+b-1}{b^2+3b+2},& 0\le x<1\\ 2x-3,& 1\le x\le 3\end{array}$. Find all possible real values of $b$ such that $f(x)$ has the smallest value at $x=1$.

Imagine you are standing on a mountain range defined by a mathematical function. You are looking for the absolute lowest point, the valley floor.

In this problem, our landscape is split into two distinct regions, joined at the coordinate $x=1$. Our goal is to ensure that this junction, $x=1$, is the absolute lowest point of the entire terrain.

Let us embark on this journey to find the values of $b$ that make this possible.

Let us first look at the right side of our landscape, where $1\le x\le 3$. Here, the function is defined as $f(x)=2x-3$.

If we plug in $x=1$, we find $f(1)=2(1)-3=-1$.

Because the slope is $2$, which is positive, this path is constantly climbing as we move to the right. This means that for any $x>1$, the value of $f(x)$ will always be greater than $-1$. The right side is already 'safe'—it will never dip below our target minimum.

Now, turn your gaze to the left, where $0\le x<1$. Here, the function is $f(x)=-x^3+C$, where $C$ is defined as:

This is a cubic function, but notice the negative sign in front of the $x^3$. The derivative is $-3x^2$, which is always non-positive.

This tells us that as we walk from $x=0$ toward $x=1$, we are constantly walking downhill. The lowest point on this segment will be the one closest to $x=1$. Mathematically, we look at the limit:

For $x=1$ to be the global minimum, the entire left branch must not dip below the value we found at the junction. We require that the 'floor' of the left branch, which is $-1+C$, must be greater than or equal to the value at the junction, $f(1)=-1$.

Setting up the inequality:

This simplifies beautifully to $C\ge 0$. The complexity of the constant $C$ suddenly collapses into a simple requirement: the rational expression must be non-negative.

Now we face the expression:

We must factorize this to see the roots. The numerator $b^3-b^2+b-1$ can be grouped as $b^2(b-1)+1(b-1)$, which gives us $(b-1)(b^2+1)$. The denominator is a classic quadratic: $(b+1)(b+2)$.

Our inequality becomes:

Since $b^2+1$ is always positive for any real $b$, we can safely divide it out, leaving us with the core of the problem:

Using the Wavy Curve Method, we plot our critical points at $b=-2,-1,1$.

Testing the intervals, we find that the expression is positive in the interval $(-2,-1)$ and in the interval $[1,\infty )$. We exclude $-2$ and $-1$ because they make the denominator zero, but we include $1$ because the numerator being zero is perfectly acceptable.

Thus, we arrive at our solution: $b\in (-2,-1)\cup [1,\infty )$.

You have successfully navigated the landscape, constrained the variables, and found the exact conditions for the minimum. This is the power of calculus—turning a complex, piecewise mystery into a clear, logical path.