Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If lim_{x - 0} [1

Q: Why can't $f(x)$ have a constant or linear term?

If $f(x)$ had a constant term $k$ or a linear term $mx$, then $\frac{f(x)}{x^2}$ would contain terms like $\frac{k}{x^2}$ or $\frac{m}{x}$. As $x \to 0$, these terms would approach infinity, making the limit undefined. Since the limit is a finite value (2), the coefficients of $x^0$ and $x^1$ must be zero.

Q: Why do we set $f'(1) = 0$ and $f'(2) = 0$?

The problem states that $f(x)$ has extreme values at $x=1$ and $x=2$. For a polynomial, extreme values (local maxima or minima) occur at stationary points where the slope of the tangent, which is the derivative $f'(x)$, is equal to zero.

Q: Is it possible for $f(x)$ to have a degree lower than 4?

The problem explicitly states that $f(x)$ is a polynomial of degree four. Therefore, we must include the $ax^4$ term. Even if $a$ turned out to be zero, we would start with the assumption of degree four to be rigorous.

Question

Let $f(x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$. If $\lim_{x 	o 0} \left[ 1 + \frac{f(x)}{x^2} ight] = 3$, then $f(2)$ is equal to

Options:

(A) 0

(B) 4

(C) -8

(D) -4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a polynomial $f(x)$ of degree four. Our goal is to determine the value of $f(2)$ by first identifying the coefficients of the polynomial.

The problem provides the limit condition:
$$\lim_{x 	o 0} \left[ 1 + \frac{f(x)}{x^2} ight] = 3$$

Subtracting 1 from both sides, we obtain:
$$\lim_{x 	o 0} \frac{f(x)}{x^2} = 2$$

### Decoding the Polynomial Structure

For the limit to exist as a finite value, $f(x)$ cannot contain constant or linear terms, as these would cause the expression to diverge as $x 	o 0$. Thus, $f(x)$ must take the form $f(x) = ax^4 + bx^3 + cx^2$.

Substituting this into our limit:
$$\lim_{x 	o 0} \frac{ax^4 + bx^3 + cx^2}{x^2} = \lim_{x 	o 0} (ax^2 + bx + c) = 2$$

As $x$ approaches 0, the terms $ax^2$ and $bx$ vanish, leaving $c = 2$. Our polynomial is now defined as:
$$f(x) = ax^4 + bx^3 + 2x^2$$

### The Power of Derivatives

We are told that $f(x)$ has extreme values at $x=1$ and $x=2$. In calculus, this implies that the derivative $f'(x)$ must be zero at these points.

First, we calculate the derivative:
$$f'(x) = 4ax^3 + 3bx^2 + 4x$$

### Solving the System

Using the conditions $f'(1) = 0$ and $f'(2) = 0$, we generate a system of linear equations:

For $x=1$:
$$4a(1)^3 + 3b(1)^2 + 4(1) = 0 \implies 4a + 3b + 4 = 0$$

For $x=2$:
$$4a(2)^3 + 3b(2)^2 + 4(2) = 0 \implies 32a + 12b + 8 = 0$$

Dividing the second equation by 4, we simplify it to $8a + 3b + 2 = 0$. We now solve the system:
1) $4a + 3b = -4$
2) $8a + 3b = -2$

Subtracting the first equation from the second yields $4a = 2$, which gives $a = \frac{1}{2}$. Substituting $a = \frac{1}{2}$ into the first equation:
$$4\left(\frac{1}{2}ight) + 3b = -4 \implies 2 + 3b = -4 \implies 3b = -6 \implies b = -2$$

### Final Calculation

The complete polynomial is:
$$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$$

To find $f(2)$, we substitute $x=2$ into the expression:
$$f(2) = \frac{1}{2}(2)^4 - 2(2)^3 + 2(2)^2$$
$$f(2) = \frac{1}{2}(16) - 2(8) + 2(4)$$
$$f(2) = 8 - 16 + 8 = 0$$

The final value is **0**.

Animated Solution for Mathematics - Differentiation: Let $f (x)$ be a polynomial of degree four having extreme values at $x = 1$ and $x = 2$ . If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is equal to

Select Answer:

Visualized Solution

Analyze the Limit Condition

Define the Polynomial $f (x)$

Find the Derivative $f^{'} (x)$

Apply Extrema Condition at $x = 1$

Apply Extrema Condition at $x = 2$

Solve for $a$

Solve for $b$

Final Polynomial Structure

Calculate $f (2)$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Polynomial Structure

The Power of Derivatives

Solving the System

Final Calculation

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Select Answer:

Visualized Solution

Analyze the Limit Condition

Define the Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Find the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Apply Extrema Condition at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Apply Extrema Condition at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Final Polynomial Structure

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(2)

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Decoding the Polynomial Structure

The Power of Derivatives

Solving the System

Final Calculation

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