Animated Solution for Mathematics - Differentiation: Let $f(x)$ be a polynomial of degree 3 such that $f(-1)=10,f(1)=-6,f(x)$ has a critical point at $x=-1$ and $f^{\prime }(x)$ has a critical point at $x=1$. Then the local minima at x = ______

A Journey into Polynomials

Welcome, my dear student. Today, we are not just solving a math problem; we are performing an autopsy on a function. We are going to reconstruct a cubic polynomial from the clues it leaves behind in the coordinate plane.

This is the essence of calculus—using the behavior of a function to uncover its very identity. Let us begin.

The Blueprint

Every cubic polynomial is a story waiting to be told. We define our protagonist as $f(x)=a{x}^3+b{x}^2+cx+d$. Here, $a,b,c,$ and $d$ are the hidden characters.

We have four unknowns, which means, mathematically, we need four independent pieces of information to solve for them. The problem provides exactly that. We are like detectives at a crime scene, gathering evidence.

Our first tool is the derivative. The derivative tells us the slope of the tangent line at any point.

But the problem goes deeper. It mentions a critical point of the derivative itself. This requires the second derivative: This is the acceleration of our curve, the rate at which the slope itself is changing.

Decoding the Clues

Let us translate the problem's conditions into the language of calculus. The problem states that $f^{\prime }(x)$ has a critical point at $x=1$.

This is a subtle but powerful hint. It means the derivative of $f^{\prime }(x)$—which is $f^{\prime \prime }(x)$—must vanish at $x=1$. Thus, $f^{\prime \prime }(1)=0$.

Substituting $x=1$ into our second derivative, we get $6a(1)+2b=0$, which simplifies beautifully to $b=-3a$. We have just reduced our complexity by one dimension!

Next, we look at the function $f(x)$ itself. It has a critical point at $x=-1$. This means the slope is zero there: $f^{\prime }(-1)=0$.

Substituting $x=-1$ into $f^{\prime }(x)=3a{x}^2+2bx+c$, we get $3a-2b+c=0$. Now, we substitute our earlier finding, $b=-3a$, into this equation.

The algebra unfolds: $3a-2(-3a)+c=0$, which leads us to $9a+c=0$, or $c=-9a$. We are on a roll!

The Algebraic Convergence

Now, we use the coordinate points. We are told $f(-1)=10$. Plugging this into our original polynomial: $a(-1{)}^3+b(-1{)}^2+c(-1)+d=10$.

This gives us $-a+b-c+d=10$. Substituting our expressions for $b$ and $c$ in terms of $a$, we get $-a-3a-(-9a)+d=10$.

Simplifying this, we find $5a+d=10$, or $d=10-5a$.

Finally, we use the last clue: $f(1)=-6$. This gives us $a+b+c+d=-6$.

Substituting everything in terms of $a$: Combining the terms, we get $-16a+10=-6$.

Solving for $a$, we find $a=1$. The mystery is unraveling! With $a=1$, we immediately find $b=-3$, $c=-9$, and $d=5$.

Our polynomial is $f(x)=x^3-3x^2-9x+5$.

The Final Verdict

We have the function, but we are not done. We need the local minima. We set $f^{\prime }(x)=3x^2-6x-9=0$.

Factoring this, we get $3(x^2-2x-3)=0$, which gives us roots at $x=3$ and $x=-1$. To distinguish between the local maxima and minima, we use the Second Derivative Test.

Our second derivative is $f^{\prime \prime }(x)=6x-6$.

At $x=-1$, $f^{\prime \prime }(-1)=-12$, which is negative, indicating a local maximum. At $x=3$, $f^{\prime \prime }(3)=12$, which is positive, indicating a local minimum.

The valley of our curve lies at $x=3$.

Take a moment to appreciate this. We started with a vague description of a curve and, through the rigorous application of calculus, we pinned it down to a specific equation. This is the power you hold as a student of physics and mathematics. Keep practicing, keep questioning, and never lose that curiosity.