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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 41

Find the cubic function f(x) = ax³ + bx² + cx + d for which ƒ( − 1) = 0, ƒ(1) = 2, ƒ(2) = 3, and ƒ(3) = 12.

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Start by writing the general form of the cubic function: \(f(x) = a x^{3} + b x^{2} + c x + d\), where \(a\), \(b\), \(c\), and \(d\) are constants to be determined.
Use the given values of the function to set up a system of equations by substituting each \(x\) value into the function and equating it to the corresponding \(f(x)\) value: - For \(f(-1) = 0\): \(a(-1)^{3} + b(-1)^{2} + c(-1) + d = 0\) - For \(f(1) = 2\): \(a(1)^{3} + b(1)^{2} + c(1) + d = 2\) - For \(f(2) = 3\): \(a(2)^{3} + b(2)^{2} + c(2) + d = 3\) - For \(f(3) = 12\): \(a(3)^{3} + b(3)^{2} + c(3) + d = 12\)
Simplify each equation to express them in terms of \(a\), \(b\), \(c\), and \(d\): - \(-a + b - c + d = 0\) - \(a + b + c + d = 2\) - \(8a + 4b + 2c + d = 3\) - \(27a + 9b + 3c + d = 12\)
Set up the system of four linear equations with four unknowns: \[\begin{cases}$ -a + b - c + d = 0 \\ a + b + c + d = 2 \\ 8a + 4b + 2c + d = 3 \\ 27a + 9b + 3c + d = 12 $\end{cases}\]
Solve this system using methods such as substitution, elimination, or matrix operations (like Gaussian elimination) to find the values of \(a\), \(b\), \(c\), and \(d\). Once found, substitute these values back into the general form to write the specific cubic function.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Cubic Functions

A cubic function is a polynomial of degree three, generally written as f(x) = ax³ + bx² + cx + d, where a, b, c, and d are constants and a ≠ 0. Understanding its general form helps in setting up equations based on given function values.
추천 영상:
4:56
Function Composition

Using Function Values to Form Equations

Given specific values of the function at certain points, you can substitute these x-values into the cubic function to create a system of equations. Each substitution yields an equation involving a, b, c, and d, which can be solved simultaneously.
추천 영상:
5:38
Verifying if Equations are Functions

Solving Systems of Linear Equations

To find the coefficients a, b, c, and d, you solve the system of linear equations formed by the function values. Techniques include substitution, elimination, or matrix methods, enabling determination of the unique cubic function fitting the given points.
추천 영상:
가이드 코스
4:27
Introduction to Systems of Linear Equations
관련 실천
교과서 질문

Perform the indicated matrix operations given that A, B and C are defined as follows. If an operation is not defined, state the reason.

A=[403501],B=[5122],C=[1111]A=\(\begin{bmatrix}\)4 & 0\\ -3 & 5\\ 0 & 1\(\end{bmatrix}\),B=\(\begin{bmatrix}\)5 & 1\\ -2 & -2\(\end{bmatrix}\),C=\(\begin{bmatrix}\)1 & -1\\ -1 & 1\(\end{bmatrix}\)

A - C

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교과서 질문

Solve the system: (Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve the system for A, B, C, and D. Then use the logarithmic equations to find w, x, y, and z.)

{2lnw+lnx+3lny2lnz=64lnw+3lnx+lnylnz=2lnw+lnx+lny+lnz=5lnw+lnxlnylnz=5\(\begin{cases}\) 2 \(\ln\) w + \(\ln\) x + 3 \(\ln\) y - 2 \(\ln\) z = -6 \\ 4 \(\ln\) w + 3 \(\ln\) x + \(\ln\) y - \(\ln\) z = -2 \\ \(\ln\) w + \(\ln\) x + \(\ln\) y + \(\ln\) z = -5 \\ \(\ln\) w + \(\ln\) x - \(\ln\) y - \(\ln\) z = 5 \(\end{cases}\)

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교과서 질문

In Exercises 39–42, find A^(-1) Check that AA^-1 = I and A^(-1)A = I

684
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교과서 질문

Find the quadratic function f(x) = ax² + bx + c for which ƒ( − 2) = −4, ƒ(1) = 2, and f(2) = 0.

874
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교과서 질문

In Exercises 37–44, use Cramer's Rule to solve each system. {x+y+z=4x2y+z=7x+3y+2z=4\(\begin{cases}\) x + y + z = 4 \\ x - 2y + z = 7 \\ x + 3y + 2z = 4 \(\end{cases}\)

864
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교과서 질문

a. Write each linear system as a matrix equation in the form AX = B. b. Solve the system using the inverse that is given for the coefficient matrix.

{wx+2y=3xy+z=4w+xy+2z=2x+y2z=4The inverse of [1120011111120112] is [0011141312120101]\(\begin{cases}\) w - x + 2y \(\quad\]\quad\) = -3 \\ \(\quad\[\quad\) x - y + z = 4 \\ -w + x - y + 2z = 2 \\ \(\quad\]\quad\) -x + y - 2z = -4 \(\end{cases}\) \\ \(\text{The inverse of }\) \(\begin{bmatrix}\) 1 & -1 & 2 & 0 \\ 0 & 1 & -1 & 1 \\ -1 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \(\end{bmatrix}\) \(\text{ is }\) \(\begin{bmatrix}\) 0 & 0 & -1 & -1 \\ 1 & 4 & 1 & 3 \\ 1 & 2 & 1 & 2 \\ 0 & -1 & 0 & -1 \(\end{bmatrix}\)

707
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