If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimu

Q: Wait, why is the derivative zero at the maximum and minimum? Can't it be undefined?

That is a fantastic conceptual question! For a polynomial, the function is smooth and differentiable everywhere. A local maximum or minimum implies the slope changes from positive to negative (or vice versa), and by Fermat's Theorem on stationary points, the derivative must be zero at these points. Since polynomials don't have cusps or discontinuities, we are guaranteed that $p'(x) = 0$ at these extrema.

Q: Why do we need the constant $B$ after integrating? Can't we just ignore it?

Never ignore the constant of integration! In indefinite integration, $B$ represents the vertical shift of the curve. Without it, you are just finding the shape of the curve, not its specific position on the $y$-axis. Since the problem gives us specific $y$-values ($8$ and $4$), $B$ is essential to pin the curve to the coordinate plane.

Q: Is there a faster way to solve this without integration?

You could assume $p(x) = ax^3 + bx^2 + cx + d$ and solve a system of four equations, but that is a trap! It is algebraically messy and prone to calculation errors. Using the derivative and integrating is the 'elegant' path—it reduces the problem to just two variables, $A$ and $B$, saving you time and keeping your math clean.

Question

If $p(x)$ be a polynomial of degree three that has a local maximum value $8$ at $x = 1$ and a local minimum value $4$ at $x = 2$; then $p(0)$ is equal to :

Options:

(A) $12$

(B) $6$

(C) $-24$

(D) $-12$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, watching a roller coaster trace the path of a cubic polynomial $p(x)$. We are given two critical landmarks: a local maximum at $(1, 8)$ and a local minimum at $(2, 4)$.

To solve this, we must listen to what the calculus tells us about the shape of the curve.

### The Calculus of Peaks and Valleys

At the local maximum and the local minimum, the roller coaster is momentarily horizontal. In the language of calculus, a horizontal tangent means the slope is zero.

Therefore, we know that the derivative of our polynomial, $p'(x)$, must be zero at $x=1$ and $x=2$.

Since $p(x)$ is a cubic polynomial, its derivative $p'(x)$ must be a quadratic. We have identified the two roots of this quadratic, allowing us to write the derivative in a powerful, factored form:

$$p'(x) = A(x-1)(x-2)$$

Here, $A$ is a scaling constant. By expanding this, we get $p'(x) = A(x^2 - 3x + 2)$.

### Reconstructing the Polynomial

We now use integration to reconstruct the original function $p(x)$ from its slope. We integrate $p'(x)$ with respect to $x$:

$$p(x) = \int A(x^2 - 3x + 2) dx$$

Performing this integration term by term, we obtain:

$$p(x) = A \left( \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right) + B$$

The constant $B$ represents the vertical shift of our curve. Since we have two specific points, $(1, 8)$ and $(2, 4)$, we can solve for both $A$ and $B$.

### Solving the System

Applying our landmarks, we first evaluate at $x=1$:

$$8 = A \left( \frac{1}{3} - \frac{3}{2} + 2 \right) + B \Rightarrow \frac{5A}{6} + B = 8$$

Next, we evaluate at $x=2$:

$$4 = A \left( \frac{8}{3} - 6 + 4 \right) + B \Rightarrow \frac{2A}{3} + B = 4$$

Subtracting the second equation from the first, the $B$ terms vanish:

$$\frac{5A}{6} - \frac{4A}{6} = 8 - 4 \Rightarrow \frac{A}{6} = 4 \Rightarrow A = 24$$

Substituting $A = 24$ back into the second equation:

$$\frac{2(24)}{3} + B = 4 \Rightarrow 16 + B = 4 \Rightarrow B = -12$$

### Final Calculation

The question asks for $p(0)$. In our integrated equation, every term contains an $x$ except for the constant $B$.

When we set $x=0$, all terms involving $x$ vanish, leaving us with $p(0) = B$.

Since we calculated $B = -12$, our final answer is **$-12$**.

Animated Solution for Mathematics - Differentiation: If $p (x)$ be a polynomial of degree three that has a local maximum value $8$ at $x = 1$ and a local minimum value $4$ at $x = 2$ ; then $p (0)$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Cubic Curve

Tangents at Critical Points

Formulating $p^{'} (x)$

Expanding the Derivative

Recovering $p (x)$ via Integration

Performing the Integration

Applying the Maximum Condition

Simplifying the First Equation

Applying the Minimum Condition

Simplifying the Second Equation

Solving for $A$

Solving for $B$

Finding $p (0)$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Calculus of Peaks and Valleys

Reconstructing the Polynomial

Solving the System

Final Calculation

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