Problem 43
Question
If \(f_{j}=\sum_{i=0}^{2} a_{i j} x^{i}, j=1,2,3\) and if \(f_{j}^{\prime} f_{j}^{\prime \prime}\) denote \(\frac{d f_{j}}{d x}, \frac{d^{2} f_{j}}{d x^{2}}\) respectively, then \(g(x)=\left|\begin{array}{lll}f_{1} & f_{2} & f_{3} \\\ f_{1}^{\prime} & f_{2}^{\prime} & f_{3}^{\prime} \\ f_{1}^{\prime \prime} & f_{2}^{\prime \prime} & f_{3}^{\prime \prime}\end{array}\right|\) is (A) a cubic in \(x\) (B) a quadratic in \(x\) (C) linear in \(x\) (D) a constant
Step-by-Step Solution
Verified Answer
(A) g(x) is a cubic in x.
1Step 1: Identify the Functions
We are given \( f_j = \sum_{i=0}^{2} a_{ij} x^{i} \) for \( j = 1, 2, 3 \). This means each \( f_j \) is a quadratic function of the form \( f_j = a_{0j} + a_{1j}x + a_{2j}x^2 \). We have three such functions: \( f_1, f_2, f_3 \).
2Step 2: Differentiate the Functions
We find the first derivative \( f_j' = \frac{d f_j}{dx} = a_{1j} + 2a_{2j}x \) and the second derivative \( f_j'' = \frac{d^2 f_j}{dx^2} = 2a_{2j} \) for each \( j = 1, 2, 3 \).
3Step 3: Set Up the Determinant
The function \( g(x) \) is defined as \[ g(x) = \left| \begin{array}{ccc} f_1 & f_2 & f_3 \ f_1' & f_2' & f_3' \ f_1'' & f_2'' & f_3'' \end{array} \right| \]. This is a 3x3 determinant with quadratic polynomials in the first row, linear polynomials in the second row, and constants in the third row.
4Step 4: Expanding the Determinant
We expand the determinant along the first row. Each element of the first row is a quadratic expression multiplied by the corresponding 2x2 determinant formed from elements of the second and third rows. This effectively results in terms of degree 3 as the highest degree, establishing that result from combining a quadratic in \( x \) with linear terms.
5Step 5: Conclusion on the Nature of g(x)
Since the determinant involves multiplying quadratic expressions with linear expressions, the highest degree term in the expansion will be \( x^3 \). Therefore, \( g(x) \) is a polynomial of degree 3.
Key Concepts
Derivative of polynomialsSecond derivativePolynomial degree
Derivative of polynomials
The derivative of a polynomial is a fundamental concept in calculus that helps us understand how functions change. If you have a polynomial, such as
For example, if \( f(x) = 3x^2 + 2x + 1 \), the derivative would be
- \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \),
- if \( f(x) = x^n \), then the derivative \( f'(x) = n x^{n-1} \).
For example, if \( f(x) = 3x^2 + 2x + 1 \), the derivative would be
- \( f'(x) = 6x + 2 \).
Second derivative
The second derivative of a function provides even deeper insights into the behavior of polynomials. Given a function \( f(x) \), the second derivative, denoted \( f''(x) \), tells us about the concavity or the "curvature" of the function.
Here's how it works:
Here's how it works:
- The first derivative \( f'(x) \) shows the slope of \( f(x) \), while the second derivative \( f''(x) \) shows how this slope changes.
- If \( f''(x) > 0 \), the function is concave up (looks like a smile), meaning it is curving upwards.
- Conversely, if \( f''(x) < 0 \), the function is concave down (frown), curving downwards.
- \( 2a \).
Polynomial degree
Understanding the degree of a polynomial is vital in recognizing the behavior and characteristics of the function. The degree of a polynomial is defined as the highest power of the variable in its expression.
- For instance, in \( 4x^3 + 3x^2 + 2 \), the highest power is 3, making it a third-degree polynomial.
- In our earlier example, as \( x \to \infty \) or \( x \to -\infty \), the graph behaves like \( x^3 \), either rising or falling without bounds.
Other exercises in this chapter
Problem 41
The value of the determinant of \(n\)th order, being given by \(\left|\begin{array}{cccc}x & 1 & 1 & \ldots \\ 1 & x & 1 & \ldots \\ 1 & 1 & x & \ldots \\ \ldot
View solution Problem 42
The value of the determinant \(\left|\begin{array}{ccc}\sqrt{x}+\sqrt{y} & 2 \sqrt{z} & \sqrt{z} \\ \sqrt{y z}+\sqrt{2 x} & z & \sqrt{2 z} \\ y+\sqrt{x z} & \sq
View solution Problem 45
\((b+c)(y+z)-a x=b-c\), \((c+a)(z+x)-b y=c-a\), \((a+b)(x+y)-c z=a-b\) where \(a+b+c \neq 0\), then \(x=\) (A) \(\frac{c-b}{a+b+c}\) (B) \(\frac{a-c}{a+b+c}\) (
View solution Problem 46
If \(x \neq 0, y \neq 0, z \neq 0\) and \(\left|\begin{array}{ccc}1+x & 1 & 1 \\\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right|=0\), then \(x^{-1}+y^{-
View solution