Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If lim (x→0) (f(x)/x²

Q: Wait, why can't $f(x)$ have a constant term or an $x$ term?

Great question! If $f(x)$ had a constant term $k$ or a linear term $mx$, the expression $\frac{f(x)}{x^2}$ would become $\frac{ax^4 + bx^3 + cx^2 + mx + k}{x^2} = ax^2 + bx + c + \frac{m}{x} + \frac{k}{x^2}$. As $x$ approaches zero, the terms $\frac{m}{x}$ and $\frac{k}{x^2}$ would approach infinity. Since the limit is given as a finite value, those terms must be zero.

Q: How do we know the degree is 4?

The problem explicitly states that $f(x)$ is a polynomial of degree 4. This tells us the highest power of $x$ is 4, which guides our general form $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. We then use the limit condition to eliminate the lower-degree terms.

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

In calculus, an extreme value (a local maximum or minimum) of a differentiable function occurs at a point where the tangent line is horizontal. A horizontal line has a slope of zero, and the slope of the tangent is given by the derivative $f'(x)$. Therefore, at any extreme point $x_0$, we must have $f'(x_0) = 0$.

Question

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $\lim_{x 	o 0} \left( \frac{f(x)}{x^2} + 1 ight) = 3$ then f(-1) is equal to :-

Options:

(A) 5/2

(B) 9/2

(C) 1/2

(D) 3/2

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Polynomials

We are exploring a fourth-degree polynomial $f(x)$ that represents a smooth, rolling landscape. The curve possesses extreme values at $x=1$ and $x=2$.

In the language of calculus, these extreme values imply that the tangent line is perfectly horizontal at these points. Mathematically, this translates to the conditions:
$$f'(1) = 0 \quad 	ext{and} \quad f'(2) = 0$$

### The Limit as a Structural Blueprint

Before analyzing the derivatives, we must evaluate the given limit:
$$\lim_{x 	o 0} \left( \frac{f(x)}{x^2} + 1 ight) = 3$$

This limit acts as a structural blueprint for the polynomial. If $f(x)$ contained a constant term or a linear term, the division by $x^2$ would cause the function to diverge to infinity as $x 	o 0$.

For the limit to be finite, $f(x)$ must take the form $f(x) = ax^4 + bx^3 + cx^2$. Substituting this into the limit expression yields:
$$\lim_{x 	o 0} (ax^2 + bx + c + 1) = 3$$

This simplifies to $c + 1 = 3$, which reveals that $c = 2$. Consequently, our function is defined as:
$$f(x) = ax^4 + bx^3 + 2x^2$$

### The Algebraic Dance

To determine the constants $a$ and $b$, we utilize the derivative conditions. Differentiating $f(x) = ax^4 + bx^3 + 2x^2$ gives:
$$f'(x) = 4ax^3 + 3bx^2 + 4x$$

Applying the conditions $f'(1) = 0$ and $f'(2) = 0$, we generate the following system of equations:
$$4a(1)^3 + 3b(1)^2 + 4(1) = 0 \implies 4a + 3b = -4$$
$$4a(2)^3 + 3b(2)^2 + 4(2) = 0 \implies 32a + 12b + 8 = 0$$

Simplifying the second equation by dividing by $4$, we obtain $8a + 3b = -2$. We now have a system of two linear equations:
$$4a + 3b = -4$$
$$8a + 3b = -2$$

Subtracting the first equation from the second yields $4a = 2$, which gives $a = \frac{1}{2}$. Substituting $a = \frac{1}{2}$ back into the first equation results in $2 + 3b = -4$, leading to $3b = -6$, or $b = -2$.

### The Final Revelation

We have successfully unlocked all the constants. The complete, exact function is:
$$f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$$

To find the value of $f(-1)$, we perform the final calculation:
$$f(-1) = \frac{1}{2}(-1)^4 - 2(-1)^3 + 2(-1)^2$$

Evaluating this expression:
$$f(-1) = \frac{1}{2}(1) - 2(-1) + 2(1) = \frac{1}{2} + 2 + 2 = \frac{9}{2}$$

The final answer is **$\frac{9}{2}$**.

Animated Solution for Mathematics - Differentiation: Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $lim_{x \to 0} (\frac{f ( x )}{x ^{2}} + 1) = 3$ then f(-1) is equal to :-

Select Answer:

Visualized Solution

Visualizing the Extrema

The Derivative at Extrema

Simplifying the Limit

Deducing the Polynomial Structure

Evaluating the Limit

Finding the Derivative

Applying Condition at $x = 1$

Applying Condition at $x = 2$

Solving for $a$ and $b$

The Final Polynomial

Calculating $f (- 1)$

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of Polynomials

The Limit as a Structural Blueprint

The Algebraic Dance

The Final Revelation

Similar Questions

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If $f (x)$ is a non-zero polynomial of degree four, having local extreme points at $x = - 1, 0, 1$ ; then the set $S = {x \in R : f (x) = f (0)}$ contains exactly :

Select Answer:

Visualized Solution

Visualizing the Extrema

The Derivative at Extrema

Simplifying the Limit

Deducing the Polynomial Structure

Evaluating the Limit

Finding the Derivative

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

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

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

The Final Polynomial

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

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of Polynomials

The Limit as a Structural Blueprint

The Algebraic Dance

The Final Revelation

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Applying Condition at $x = 2$

Solving for $a$ and $b$

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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

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Let $f : R \to R$ be a polynomial function of degree four having extreme values at $x = 4$ and $x = 5$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{2}} = 5$ , then $f (2)$ is equal to :

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4$ , then which one of the following is not true?

Let $f (x)$ be a polynomial of degree $6$ in $x$ , in which the coefficient of $x^{6}$ is unity and it has extrema at $x = - 1$ and $x = 1$ . If $lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 1$ , then $5 \cdot f (2)$ is equal to

If $f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}$ , for every real number $x$ , then the minimum value of $f$

If $f (x)$ is a non-zero polynomial of degree four, having local extreme points at $x = - 1, 0, 1$ ; then the set $S = {x \in R : f (x) = f (0)}$ contains exactly :