Animated Solution for Mathematics - Differentiation: The sum of all the local minimum values of the twice differentiable function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=x^3-3x^2-\frac{3f^{\prime \prime }(2)}{2}x+f^{\prime \prime }(1)$ is :

Imagine you are standing before a locked door. The lock is complex, filled with gears and levers that seem to move on their own. This is exactly how the function

feels at first glance. It is a self-referential puzzle.

It defines itself using its own second derivative. But here is the secret that separates the novice from the master: do not be intimidated by the notation.

In the world of calculus, $f^{\prime \prime }(2)$ and $f^{\prime \prime }(1)$ are not variables; they are simply numbers waiting to be discovered. They are the constants that hold the structure of our polynomial together.

The Detective Work

To solve this, we must act like detectives. We need to find the values of $f^{\prime \prime }(2)$ and $f^{\prime \prime }(1)$.

How do we do that? We differentiate. Let us take our function $f(x)$ and find its first derivative, $f^{\prime }(x)$.

Differentiating $x^3$ gives us $3x^2$, and $-3x^2$ becomes $-6x$. The term involving $x$ simply leaves us with its coefficient, $-\frac{3f^{\prime \prime }(2)}{2}$.

And the term $f^{\prime \prime }(1)$? It is a constant, so its derivative is zero. We are left with:

Now, we need the second derivative, $f^{\prime \prime }(x)$. Differentiating again, the $3x^2$ becomes $6x$, the $-6x$ becomes $-6$, and our constant term vanishes entirely.

Suddenly, the fog clears. We find that:

This is the key that unlocks the entire problem!

The Reveal

With $f^{\prime \prime }(x)=6x-6$ in our hands, the mystery constants are no longer mysterious. We simply plug in the values.

For $f^{\prime \prime }(2)$, we calculate $6(2)-6$, which gives us $6$. For $f^{\prime \prime }(1)$, we calculate $6(1)-6$, which gives us $0$.

Just like that, the skeleton of our function is revealed. Substituting these back into our original equation, we get:

Simplifying this, we arrive at the beautiful, concrete polynomial:

The Calculus of Peaks and Valleys

Now that we have the function, we are on familiar ground. To find the local minimum, we need to find the critical points where the slope is zero.

We set $f^{\prime }(x)=3x^2-6x-9=0$. Dividing by $3$, we get $x^2-2x-3=0$.

Factoring this quadratic, we find $(x-3)(x+1)=0$. Our critical points are $x=3$ and $x=-1$.

Which one is the minimum? We use the Second Derivative Test.

We know $f^{\prime \prime }(x)=6x-6$. At $x=-1$, $f^{\prime \prime }(-1)=-12$, which is negative, indicating a local maximum (a peak).

At $x=3$, $f^{\prime \prime }(3)=12$, which is positive, indicating a local minimum (a valley). We have found our target!

The Final Calculation

Finally, we must answer the question asked. We need the local minimum value.

We plug $x=3$ back into our function:

This simplifies to $27-27-27$, which equals $-27$.

The journey is complete. We started with a confusing, self-referential expression and ended with a clear, definitive value of $-27$. This is the power of calculus—taking the unknown and, through systematic logic, making it known.