Problem 73
Question
Find the zeros (if any) of the rational function. $$f(x)=5-\frac{3}{x-2}$$
Step-by-Step Solution
Verified Answer
The zero of the function \(f(x)=5-\frac{3}{x-2}\) is \(x = \frac{13}{5}\)
1Step 1: Set the function equal to zero
Here, take the function \(f(x) = 5 - \frac{3}{x-2}\) and set it equal to zero: \[0 = 5- \frac{3}{x-2}\].
2Step 2: Solve the equation
Next, solve the equation for \(x\). First, move the term \(-\frac{3}{x-2}\) to the other side of the equation: \[\frac{3}{x-2} = 5\]. Then, make the term \(x-2\) the subject of the formula by multiplying both sides by it: \[3 = 5(x - 2)\]. Finally, solve for \(x\) by distributing the 5 on the right side and isolating \(x\): \[3 = 5x - 10\], \[13 = 5x\], \(x = \frac{13}{5}\].
3Step 3: Checking the Solution
Lastly, ensure that the solution doesn't result in the denominator of the original function being zero. In this case, \(x-2\) can't be equal to zero or \(x\) can't be 2. The solution \(x = \frac{13}{5}\), which is not 2, is thus a valid solution.
Key Concepts
Understanding a Rational FunctionSolving Rational EquationsAlgebraic Manipulation in Rational Equations
Understanding a Rational Function
A rational function is a type of function which is represented by the ratio of two polynomials. In its simplest form, a rational function is expressed as \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). It's important to understand that the domain of a rational function excludes any real number that causes the denominator to be zero since division by zero is undefined.
To better grasp the concept, let's consider our example \( f(x) = 5 - \frac{3}{x-2} \). Here, \(5\) is a constant polynomial, and the term \(-\frac{3}{x-2}\) constitutes the rational part where the numerator is \(-3\) and the denominator is the binomial \(x-2\). The presence of the variable in the denominator is a clear identifier of a rational function.
To better grasp the concept, let's consider our example \( f(x) = 5 - \frac{3}{x-2} \). Here, \(5\) is a constant polynomial, and the term \(-\frac{3}{x-2}\) constitutes the rational part where the numerator is \(-3\) and the denominator is the binomial \(x-2\). The presence of the variable in the denominator is a clear identifier of a rational function.
Solving Rational Equations
The process of solving rational equations involves finding the value of the variable that makes the equation true. To solve for the zeros of a rational function, we set the entire function equal to zero and solve for the variable. When solving rational equations, there are several key steps to follow, such as clearing the fractions, isolating the variable, and, crucially, checking that the solution doesn't make the denominator zero.
In the provided exercise, we set \( f(x) = 0 \) to solve for the zeros of the rational function given. This involves algebraic manipulation to clear the fraction and isolate \( x \). The solution process also requires you to be mindful of any restrictions on the values of \( x \), which come from the original equation's denominator.
In the provided exercise, we set \( f(x) = 0 \) to solve for the zeros of the rational function given. This involves algebraic manipulation to clear the fraction and isolate \( x \). The solution process also requires you to be mindful of any restrictions on the values of \( x \), which come from the original equation's denominator.
Algebraic Manipulation in Rational Equations
Understanding and employing algebraic manipulation is essential when working with rational functions. Algebraic manipulation includes operations like addition, subtraction, multiplication, division, and factorization, among others, to rearrange equations and expressions to make them solvable. It is a powerful toolset that allows mathematicians and students alike to simplify and solve complex problems.
Regarding our exercise, algebraic manipulation first involves moving all terms involving \( x \) to one side of the equation and constants to the other. This is done to isolate \( x \) and solve for it effectively. The step of multiplying both sides of the equation by \( x - 2 \) is a critical example of algebraic manipulation, intended to eliminate the fraction. After simplifying, we solve for \( x \) as a numerical value, ensuring that our algebraic maneuvers do not violate any mathematical principles, such as dividing by zero.
Regarding our exercise, algebraic manipulation first involves moving all terms involving \( x \) to one side of the equation and constants to the other. This is done to isolate \( x \) and solve for it effectively. The step of multiplying both sides of the equation by \( x - 2 \) is a critical example of algebraic manipulation, intended to eliminate the fraction. After simplifying, we solve for \( x \) as a numerical value, ensuring that our algebraic maneuvers do not violate any mathematical principles, such as dividing by zero.
Other exercises in this chapter
Problem 72
Determine whether the sequence is arithmetic, geometric, or neither. $$66,55,44,33,22, \dots$$
View solution Problem 73
Check for symmetry with respect to both axes and to the origin. Then determine whether the function is even, odd, or neither. $$y=e^{x}$$
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Find the vertex, focus, and directrix of the parabola and sketch its graph. Use a graphing utility to verify your graph. $$x^{2}+4 x+6 y-2=0$$
View solution Problem 73
Find the equation of the hyperbola for any point at which the difference between its distances from the points (2,2) and (10,2) is 6
View solution