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Chapter 1: Problem 119

If \(F(x)=\frac{C-x}{D-x}, F(-2)\) is undefined, and \(F(-1)=4,\) find \(C\) and \(D\)

Short Answer

Expert verified
C = -5, D = -2.

Step by step solution

01

Understand the Undefined Expression

The function \(F(x) = \frac{C-x}{D-x}\) is undefined when the denominator is zero. We need \(D - x = 0\) to make the function undefined. Given \(F(-2)\) is undefined, set \(D - (-2) = 0\). Solve this equation to find \(D = -2\).
02

Use the Known Function Value

We know that \(F(-1) = 4\). Substitute \(x = -1\) into the function and set it equal to 4: \(F(-1) = \frac{C - (-1)}{D - (-1)} = 4\). Simplify to \(\frac{C + 1}{D + 1} = 4\).
03

Use the Value of D

Substitute \(D = -2\) from Step 1 into the equation from Step 2: \(\frac{C + 1}{-2 + 1} = 4\). Simplify the denominator to get \(\frac{C + 1}{-1} = 4\).
04

Solve for C

Simplify \(\frac{C + 1}{-1} = 4\) to \(C + 1 = -4\). Solve for \(C\), giving \(C = -4 - 1 = -5\).

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Key Concepts

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

Undefined Expressions
Rational functions can often become undefined if the denominator equals zero. This is a common scenario because division by zero is not defined in mathematics. In our exercise, the function is given as \( F(x) = \frac{C-x}{D-x} \). For this function to become undefined, we must ensure that the denominator \( D-x \) equals zero.

Given that \( F(-2) \) is undefined, it means when we substitute \( x = -2 \) into the function, the denominator resolves to zero: \( D - (-2) = 0 \).
Solving this, we find:
  • \( D + 2 = 0 \)
  • \( D = -2 \)
It is crucial to remember that an undefined expression often points directly to a restriction in the domain of the function, specifically where the denominator becomes zero.
Solving Equations
The exercise also requires solving an equation tied to a known function value. We know from the problem statement that \(F(-1) = 4\). Thus, we substitute \(x = -1\) into the function to form an equation:
\( \frac{C - (-1)}{D - (-1)} = 4 \).
This simplifies to:
  • \(\frac{C + 1}{D + 1} = 4\)
The next step involves replacing \(D\) with its found value from the undefined expression section, \(D = -2\). It changes our equation to \(\frac{C + 1}{-2 + 1} = 4\).
This simplifies further to:
  • \(\frac{C + 1}{-1} = 4\)
By solving this, you ensure the problem is consistent with all given values and confirmations.
Function Evaluation
Evaluating a function involves substituting specific values into the variable \(x\) and simplifying. In this problem, we already determined that \(F(-2)\) leads to an undefined expression and verified that \(F(-1)\) holds a value.

Using the known function evaluation \(F(-1) = 4\), substitute and simplify the expression \(\frac{C + 1}{-1} = 4\) from the solving equations section:
  • Multiply both sides by \(-1\) to clear the fraction, resulting in \(C + 1 = -4\).
  • Then solve for \(C\) to find \(C = -5\).
This leads us to successfully evaluate the function and confirm the values of unknowns \(C\) and \(D\), which are often the desired outcome when dealing with rational functions. Function evaluation helps us connect expressions with real-world quantities, validating calculations.

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Most popular questions from this chapter

In Exercises \(77-81,\) for the functions \(f(x)=x+2\) and \(g(x)=x^{2}-4,\) find the indicated function and state its domain. Explain the mistake that is made in each problem. $$\frac{g}{f}$$ Solution: $$\begin{aligned}\frac{g(x)}{f(x)} &=\frac{x^{2}-4}{x+2} \\\&=\frac{(x-2)(x+2)}{x+2} \\\&=x-2\end{aligned}$$ Domain: \((-\infty, \infty)\) This is incorrect. What mistake was made?

In Exercises \(51-60\), show that \(f(g(x))=x\) and \(g(f(x))=x\). $$f(x)=\sqrt{25-x^{2}}, g(x)=\sqrt{25-x^{2}} \text { for } 0 \leq x \leq 5$$

Determine the domain of the function \(g(t)=\sqrt{3-t}\) and express it in interval notation. Solution: What can \(t\) be? Any nonnegative real number. \(3-t>0\) \(3>t \quad\) or \(\quad t<3\) Domain: \((-\infty, 3)\) This is incorrect. What mistake was made?

In Exercises \(83-86,\) determine whether each statement is true or false. For any functions \(f\) and \(g,(f \circ g)(x)\) exists for all values of \(x\) that are in the domain of \(g(x),\) provided the range of \(g\) is a subset of the domain of \(f\).

In Exercises \(39-50\), evaluate \(f(g(1))\) and \(g(f(2)),\) if possible. $$f(x)=\sqrt[3]{x-3}, \quad g(x)=\frac{1}{x-3}$$
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