Animated Solution for Mathematics - Differentiation: Let $f:R\to R$ be a function such that $f(x)=x^3+x^2{f}^{\prime }(1)+x{f}^{\prime \prime }(2)+f^{\prime \prime \prime }(3),x\in R$. Then $f(2)$ equal :

Welcome, future engineers! Today, we are going to dismantle a problem that often terrifies students at first glance.

You see an expression like $f(x)=x^3+x^2{f}^{\prime }(1)+x{f}^{\prime \prime }(2)+f^{\prime \prime \prime }(3)$ and your brain immediately screams, "How can a function be defined by its own derivatives?"

But here is the secret: this is not a monster; it is a puzzle. The key to mastering JEE-level calculus is to look past the intimidating notation and identify the hidden simplicity.

The Transformation

The first step is to realize that $f^{\prime }(1)$, $f^{\prime \prime }(2)$, and $f^{\prime \prime \prime }(3)$ are not variables. They are constants.

When you evaluate a derivative at a specific point, you get a number. So, let us stop being intimidated by the notation and assign them simple labels: let $A=f^{\prime }(1)$, $B=f^{\prime \prime }(2)$, and $C=f^{\prime \prime \prime }(3)$.

Now, our function transforms into a friendly, standard cubic polynomial:

Suddenly, the problem feels much more manageable, doesn't it?

The Calculus Engine

Now that we have our polynomial form, we need to find the values of $A$, $B$, and $C$. To do this, we need to express the derivatives of our function.

Let us differentiate $f(x)=x^3+A{x}^2+Bx+C$ with respect to $x$. Using the power rule, we get the first derivative:

Differentiating again gives us the second derivative:

Finally, one more differentiation gives us the third derivative:

Notice how the constants $A$ and $B$ vanish as we take higher derivatives? This is the beauty of polynomials!

The System of Equations

Now, we use our definitions of $A$, $B$, and $C$ to build our system of equations. We know $C=f^{\prime \prime \prime }(3)$.

Since we found $f^{\prime \prime \prime }(x)=6$ for any $x$, it follows that $C=6$.

Next, we use $A=f^{\prime }(1)$. Substituting $x=1$ into our expression for $f^{\prime }(x)$, we get:

This simplifies to $A=3+2A+B$, or:

This is our first equation. Then, we use $B=f^{\prime \prime }(2)$. Substituting $x=2$ into our expression for $f^{\prime \prime }(x)$, we get:

This simplifies to $B=12+2A$, or:

This is our second equation.

The Final Reveal

We now have a simple system of linear equations: $A+B=-3$ and $2A-B=-12$.

Adding these two equations together, the $B$ terms cancel out perfectly:

This gives $3A=-15$, so $A=-5$. Substituting $A=-5$ back into $A+B=-3$, we get $-5+B=-3$, which means $B=2$.

We have all our constants: $A=-5$, $B=2$, and $C=6$. Our function is fully revealed:

Finally, calculating $f(2)$ is just a matter of careful substitution:

And there you have it! We have conquered the monster by breaking it down into simple, logical steps. The final answer is -2.

Keep this mindset, and you will find that even the most complex JEE problems are just puzzles waiting to be solved.