Animated Solution for Mathematics - Differentiation: Let $f(x)$ be a cubic polynomial with $f(1)=-10,f(-1)=6$, and has a local minima at $x=1$, and $f^{\prime }(x)$ has a local minima at $x=-1$. Then $f(3)$ is equal to .

Unveiling the Hidden Structure

Imagine you are an architect, but instead of steel and glass, you are building with the elegant curves of calculus. We are given a cubic polynomial $f(x)$, a shape that can twist and turn, but is governed by strict mathematical laws.

Our mission is to reconstruct this polynomial from a few scattered clues: its values at specific points and the locations of its extrema. This is not just algebra; it is a detective story where every derivative is a fingerprint.

The Second Derivative as a Compass

We start with the knowledge that $f(x)$ is a cubic. This tells us everything about its "DNA." If $f(x)$ is a cubic, then its first derivative $f^{\prime }(x)$ must be a quadratic, and its second derivative $f^{\prime \prime }(x)$ must be a linear function.

The problem drops a crucial hint: $f^{\prime }(x)$ has a local minimum at $x=-1$. In the language of calculus, if a function has a local minimum, its derivative must vanish at that point. Therefore, the derivative of $f^{\prime }(x)$, which is $f^{\prime \prime }(x)$, must be zero at $x=-1$.

This gives us our first anchor: $f^{\prime \prime }(-1)=0$. Since $f^{\prime \prime }(x)$ is linear and vanishes at $x=-1$, it must take the form $f^{\prime \prime }(x)=k(x+1)$. To make our integration journey smoother, we choose $k=6a$, giving us the elegant expression:

The Ascent through Integration

Now, we climb back up the ladder of derivatives. To find $f^{\prime }(x)$, we integrate $f^{\prime \prime }(x)$ with respect to $x$:

Here, $b$ is our first constant of integration. We are told $f(x)$ has a local minimum at $x=1$. This means the slope of the tangent at $x=1$ is zero, or $f^{\prime }(1)=0$.

Substituting this into our expression for $f^{\prime }(x)$:

With $b$ defined in terms of $a$, our derivative becomes a unified expression: $f^{\prime }(x)=3a(x+1{)}^2-12a$. The structure is beginning to emerge.

Reconstructing the Polynomial

We integrate one last time to reach the original function $f(x)$:

Now, we have two unknowns, $a$ and $c$, and two remaining clues: $f(1)=-10$ and $f(-1)=6$. Let us apply them like pieces of a puzzle. First, at $x=1$:

Next, at $x=-1$:

The Final Resolution

We are left with a simple system of linear equations. Subtracting the first from the second, we find:

With $a=1$, we quickly find $c$ by substituting back into $12a+c=6$:

Our cubic polynomial is now fully revealed:

Finally, we calculate $f(3)$ to complete our journey:

Look at what we have achieved. We didn't just solve for a number; we reconstructed a mathematical object from its properties. This is the beauty of calculus—it allows us to see the "skeleton" of a function through its derivatives. The final answer is 22.