Let J ( X ) = 3 4 X + 2 3 J(x)=\frac{3}{4} X+\frac{2}{3} J ( X ) = 4 3 ​ X + 3 2 ​ . Find J ( − 2 J(-2 J ( − 2 ].Select The Correct Response:A. 17 12 \frac{17}{12} 12 17 ​ B. − 5 6 -\frac{5}{6} − 6 5 ​ C. 2 3 \frac{2}{3} 3 2 ​ D. − 1 6 -\frac{1}{6} − 6 1 ​

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Introduction

In this article, we will be solving for the value of j(-2) given the function j(x) = (3/4)x + (2/3). We will use algebraic manipulation to find the correct answer.

Understanding the Function

The function j(x) is a linear function, which means it can be written in the form j(x) = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is 3/4 and the y-intercept is 2/3.

Substituting x = -2

To find the value of j(-2), we need to substitute x = -2 into the function j(x) = (3/4)x + (2/3).

j(-2) = (3/4)(-2) + (2/3)

Simplifying the Expression

To simplify the expression, we need to multiply the numbers and then add them together.

j(-2) = -3/2 + 2/3

Finding a Common Denominator

To add the fractions, we need to find a common denominator. The least common multiple of 2 and 3 is 6.

j(-2) = (-3/2)(3/3) + (2/3)(2/2)

j(-2) = -9/6 + 4/6

Adding the Fractions

Now that we have a common denominator, we can add the fractions.

j(-2) = (-9 + 4)/6

j(-2) = -5/6

Conclusion

Therefore, the value of j(-2) is -5/6.

Answer

The correct answer is B. -5/6.

Discussion

This problem requires the use of algebraic manipulation to find the value of j(-2). The student needs to substitute x = -2 into the function, simplify the expression, find a common denominator, and add the fractions. This problem requires a strong understanding of algebraic concepts and the ability to apply them to solve a problem.

Key Takeaways

  • The function j(x) = (3/4)x + (2/3) is a linear function.
  • To find the value of j(-2), we need to substitute x = -2 into the function.
  • We need to simplify the expression, find a common denominator, and add the fractions.
  • The value of j(-2) is -5/6.

Related Topics

  • Linear functions
  • Algebraic manipulation
  • Finding a common denominator
  • Adding fractions

Practice Problems

  1. Find the value of j(3) given the function j(x) = (2/3)x + (1/2).
  2. Find the value of j(-1) given the function j(x) = (1/2)x + (3/4).
  3. Find the value of j(2) given the function j(x) = (3/4)x + (2/3).

Solutions

  1. j(3) = (2/3)(3) + (1/2) = 2 + 1/2 = 5/2
  2. j(-1) = (1/2)(-1) + (3/4) = -1/2 + 3/4 = -2/4 + 3/4 = 1/4
  3. j(2) = (3/4)(2) + (2/3) = 3/2 + 2/3 = (9 + 4)/6 = 13/6
    Q&A: Let's Solve for j(x) =============================

Introduction

In our previous article, we solved for the value of j(-2) given the function j(x) = (3/4)x + (2/3). In this article, we will answer some frequently asked questions related to the function j(x).

Q: What is the domain of the function j(x)?

A: The domain of the function j(x) is all real numbers, since the function is defined for any value of x.

Q: What is the range of the function j(x)?

A: The range of the function j(x) is all real numbers, since the function can take on any value.

Q: Is the function j(x) a linear function?

A: Yes, the function j(x) is a linear function, since it can be written in the form j(x) = mx + b, where m is the slope and b is the y-intercept.

Q: What is the slope of the function j(x)?

A: The slope of the function j(x) is 3/4.

Q: What is the y-intercept of the function j(x)?

A: The y-intercept of the function j(x) is 2/3.

Q: How do I find the value of j(x) for a given value of x?

A: To find the value of j(x) for a given value of x, you need to substitute x into the function j(x) = (3/4)x + (2/3) and simplify the expression.

Q: Can I use the function j(x) to model real-world situations?

A: Yes, the function j(x) can be used to model real-world situations, such as the cost of a product or the temperature of a room.

Q: How do I graph the function j(x)?

A: To graph the function j(x), you need to plot the points (x, j(x)) for a range of values of x and connect the points with a line.

Q: Can I use the function j(x) to solve systems of equations?

A: Yes, the function j(x) can be used to solve systems of equations, such as the system of equations:

j(x) = (3/4)x + (2/3) j(x) = (2/3)x + (1/2)

Q: How do I find the inverse of the function j(x)?

A: To find the inverse of the function j(x), you need to swap the x and y variables and solve for y.

Q: Can I use the function j(x) to model periodic phenomena?

A: No, the function j(x) is not suitable for modeling periodic phenomena, since it is a linear function.

Q: How do I use the function j(x) to model exponential growth or decay?

A: No, the function j(x) is not suitable for modeling exponential growth or decay, since it is a linear function.

Conclusion

In this article, we answered some frequently asked questions related to the function j(x). We hope that this article has been helpful in understanding the function j(x) and its applications.

Key Takeaways

  • The domain of the function j(x) is all real numbers.
  • The range of the function j(x) is all real numbers.
  • The function j(x) is a linear function.
  • The slope of the function j(x) is 3/4.
  • The y-intercept of the function j(x) is 2/3.
  • The function j(x) can be used to model real-world situations.
  • The function j(x) can be used to solve systems of equations.
  • The function j(x) can be used to find the inverse of a function.

Related Topics

  • Linear functions
  • Algebraic manipulation
  • Finding the inverse of a function
  • Solving systems of equations
  • Modeling real-world situations

Practice Problems

  1. Find the value of j(4) given the function j(x) = (2/3)x + (1/2).
  2. Find the value of j(-3) given the function j(x) = (3/4)x + (2/3).
  3. Find the inverse of the function j(x) = (3/4)x + (2/3).

Solutions

  1. j(4) = (2/3)(4) + (1/2) = 8/3 + 1/2 = (16 + 3)/6 = 19/6
  2. j(-3) = (3/4)(-3) + (2/3) = -9/4 + 2/3 = (-27 + 8)/12 = -19/12
  3. The inverse of the function j(x) = (3/4)x + (2/3) is j^(-1)(x) = (4/3)x - (2/3).