Let x = 2 be a local minima of the function f(x) = 2x⁴ - 18x² + 8x + 12, x ∈ (-4, 4). If M is local

Q: Wait, why can't I just plug the value of $x$ directly into the original function? It seems like standard procedure.

Ah, the classic trap! While technically correct, plugging a value like $x = \frac{\sqrt{6}-2}{2}$ into a quartic polynomial $2x^4 - 18x^2 + 8x + 12$ is an algebraic nightmare waiting to happen. You will likely make a calculation error. The Division Algorithm is your shield against such errors—it turns a complex evaluation into a simple linear one.

Q: How do I know which root of the quadratic equation corresponds to the local maximum?

Excellent question. We have two roots from $2x^2 + 4x - 1 = 0$. We use the Second Derivative Test. Calculate $f''(x) = 24x^2 - 36$. If $f''(x) 0$, it is a local minimum. Always verify!

Q: Why is $(x-2)$ a factor of the derivative $f'(x)$?

Because $x=2$ is given as a local minimum. In calculus, a local minimum (or maximum) must be a stationary point where the slope of the tangent is zero. Therefore, $f'(2)$ must be $0$. By the Factor Theorem, if $f'(2) = 0$, then $(x-2)$ must be a factor of the polynomial $f'(x)$.

Question

Let $x = 2$ be a local minima of the function $f(x) = 2x^4 - 18x^2 + 8x + 12, x \in (-4, 4)$. If $M$ is local maximum value of the function $f$ in $(-4, 4)$, then $M =$

Options:

(A) $12\sqrt{6} - \frac{33}{2}$

(B) $12\sqrt{6} - \frac{31}{2}$

(C) $18\sqrt{6} - \frac{33}{2}$

(D) $18\sqrt{6} - \frac{31}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are examining the quartic function defined by:

$$f(x) = 2x^4 - 18x^2 + 8x + 12$$

We are given that $x=2$ is a local minimum. In calculus, a local minimum occurs at a stationary point where the first derivative vanishes. Thus, we must have $f'(2) = 0$.

### The Derivative and Factorization

To identify all critical points, we first compute the derivative of $f(x)$:

$$f'(x) = 8x^3 - 36x + 8$$

Since $x=2$ is a root of $f'(x) = 0$, the term $(x-2)$ must be a factor of the derivative. Factoring out a constant $4$, we obtain:

$$f'(x) = 4(2x^3 - 9x + 2)$$

By performing polynomial division, we factor the cubic expression:

$$2x^3 - 9x + 2 = (x-2)(2x^2 + 4x - 1)$$

The remaining critical points are the roots of the quadratic equation $2x^2 + 4x - 1 = 0$. Applying the quadratic formula, we find:

$$x = \frac{-4 \pm \sqrt{16 - 4(2)(-1)}}{2(2)} = \frac{-4 \pm \sqrt{24}}{4} = \frac{-2 \pm \sqrt{6}}{2}$$

### The Second Derivative Test

We must determine which of these candidates corresponds to the local maximum. We compute the second derivative:

$$f''(x) = 24x^2 - 36$$

We test the candidate $x_1 = \frac{\sqrt{6}-2}{2}$. Substituting this into $f''(x)$ yields a negative value, which confirms that $x_1$ is indeed the location of the **local maximum**.

### The Division Algorithm Trick

To evaluate $f(x_1)$ without tedious substitution, we use the Division Algorithm. Since $2x^2 + 4x - 1 = 0$ at our critical point, we divide $f(x)$ by this quadratic:

$$f(x) = (2x^2 + 4x - 1)(x^2 - 2x - \frac{9}{2}) + (24x + \frac{15}{2})$$

Because the first term vanishes at $x_1$, the function value simplifies to the remainder:

$$M = 24x_1 + \frac{15}{2}$$

Substituting $x_1 = \frac{\sqrt{6}-2}{2}$ into the remainder expression:

$$M = 24\left(\frac{\sqrt{6}-2}{2}\right) + \frac{15}{2} = 12(\sqrt{6}-2) + \frac{15}{2}$$

### Final Calculation

Simplifying the expression for $M$:

$$M = 12\sqrt{6} - 24 + \frac{15}{2} = 12\sqrt{6} - \frac{48}{2} + \frac{15}{2}$$

The final value of the local maximum is:

$$\mathbf{M = 12\sqrt{6} - \frac{33}{2}}$$

Animated Solution for Mathematics - Differentiation: Let $x = 2$ be a local minima of the function $f (x) = 2 x^{4} - 18 x^{2} + 8 x + 12, x \in (- 4, 4)$ . If $M$ is local maximum value of the function $f$ in $(- 4, 4)$ , then $M =$

Select Answer:

Visualized Solution

Analyzing the Given Function

Finding the First Derivative

Calculating $f^{'} (x)$

Identifying Critical Points

Polynomial Division

Solving the Quadratic Equation

The Second Derivative Test

Testing the Critical Points

Simplifying the Evaluation of $M$

Applying the Division Algorithm

Substituting $x_{1}$

Final Calculation of $M$

Concluding the Answer

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Derivative and Factorization

The Second Derivative Test

The Division Algorithm Trick

Final Calculation

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Visualized Solution

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Applying the Division Algorithm

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The Second Derivative Test

The Division Algorithm Trick

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