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Chapter 1: Problem 40

Find the value of \(x\) in the domain of \(f(x)=4-3 x\) for which \(f(x)=7\).

Short Answer

Expert verified
Therefore, the value of \(x\) for which \(f(x) = 7\) in the given function is \(x = -1\).

Step by step solution

01

Identify the Equation

The first step is to set the function \(f(x) = 4 - 3x\) equal to the given value of 7. This gives us the equation \(4 - 3x = 7\).
02

Rearranging the Equation

The next step is to solve for \(x\). We can begin by subtracting 4 from both sides of the equation to isolate the term containing \(x\) on one side. This gives: \(-3x = 7 - 4\) or \(-3x = 3\).
03

Solving for \(x\)

Now, we divide both sides of the equation by -3 to find the value of \(x\). This gives us: \(x = \frac{3}{-3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
Solving equations is a fundamental skill in algebra. In the given exercise, we start with a linear equation derived from a function, which is a straightforward process but requires careful handling of arithmetic operations.
To solve the equation:
  • First, we identified the equation by setting the function equal to a specific value, leading us to the linear equation \(4 - 3x = 7\).
  • Then, we rearranged the equation to isolate the variable. We did this by subtracting 4 from both sides, obtaining \(-3x = 3\).
  • Finally, we solved for \(x\) by dividing both sides by -3, which resulted in \(x = \frac{3}{-3}\). This simplifies to \(x = -1\).
These steps highlight **rearranging** equations to isolate the variable. Practice and familiarity with these methods enable confident problem-solving across many types of algebraic equations, making it crucial for students to master.
Function Evaluation
Function evaluation can seem complex, but it's a very structured process. It involves substituting a value into the function and performing the necessary calculations.
With our function \(f(x) = 4 - 3x\), when we evaluated it for specific values to confirm the consistency of our findings, we realized:
  • To check the solution \(x = -1\), substitute \(-1\) back into the function: \(f(-1) = 4 - 3(-1) = 4 + 3 = 7\).
  • This confirms that when \(x = -1\), \(f(x)\) meets the desired function value of 7.
Evaluation ensures accuracy of derived solutions and affirms understanding of how a function behaves. Through practice, students will gain confidence, being able to determine the output of any function for any input efficiently.
Domain of a Function
Understanding the domain of a function is crucial because it defines all the possible inputs that the function can accept.
For the function \(f(x) = 4 - 3x\), we find:
  • This is a linear function, which typically has a domain of all real numbers. There are no restrictions (like denominators or square roots) that limit the inputs.
  • The domain \(D\) is expressed as \(\{ x \mid x \in \mathbb{R} \}\), meaning any real number can be substituted for \(x\).
Knowing the domain is crucial as it informs you which values are acceptable for both solving equations and evaluating functions. It is an essential part of understanding and working with functions and ensures that the operations you perform are valid within the realm of the function's application.

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Most popular questions from this chapter

Sketch the graph of the piecewisedefined function. $$s(x)=\left\\{\begin{array}{ll} 1 & \text { if } x \text { is an integer } \\ 2 & \text { if } x \text { is not an integer } \end{array}\right.$$

Use interval notation to express the solution set of each inequality. $$|x+3| \geq 5$$

The notation \(\left.f(x)\right|_{a} ^{b}\) is used to denote the difference \(f(b)-f(a) .\) That is, $$\left.f(x)\right|_{a} ^{b}=f(b)-f(a)$$ Evaluate \(\left.f(x)\right|_{0} ^{b}\) for the given function \(f\) and the indicated values of \(a\) and \(b\). $$f(x)=2 x^{3}-3 x^{2}-x ;\left.f(x)\right|_{0} ^{2}$$

Determine whether 0 is in the range of \(g(x)=\frac{1}{x-3}\)

Solve by completing the square or by using the quadratic formula. $$\sqrt{2} x^{2}+3 x+\sqrt{2}-0$$
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