Problem 11
Question
In Problems 9-12, determine at which points \(f(x)\) is discontinuous. $$ f(x)=\left\\{\begin{array}{cl} \frac{x^{2}-3 x+2}{x-2} & \text { if } x \neq 1 \\ 1 & \text { if } x=1 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The function is discontinuous at \( x = 1 \) and \( x = 2 \).
1Step 1: Identify the function parts
The function \( f(x) \) is defined piecewise with two parts: \( \frac{x^2 - 3x + 2}{x - 2} \) for \( x eq 1 \) and 1 for \( x = 1 \). Thus, we need to check for discontinuity at \( x = 1 \) and any possible undefined points in \( \frac{x^2 - 3x + 2}{x - 2} \).
2Step 2: Analyze the quadratic expression
The expression \( x^2 - 3x + 2 \) can be factored into \( (x-1)(x-2) \). Hence, the function becomes \( \frac{(x-1)(x-2)}{x-2} \) when \( x eq 1 \).
3Step 3: Simplify the function and determine undefined points
After canceling \( x-2 \) (since \( x eq 2 \)), \( f(x) = x-1 \) for \( x eq 2 \). The function is undefined at \( x = 2 \), because of division by zero.
4Step 4: Check for discontinuity at \( x = 1 \)
As \( x \to 1 \), \( f(x) = x-1 \to 0 \). The function value at \( x = 1 \) is 1, thus \( f(x) \) is discontinuous at \( x = 1 \) because the limit and the function value at \( x = 1 \) are not equal.
5Step 5: Confirm discontinuity at \( x = 2 \)
Since \( f(x) \) is undefined at \( x = 2 \), it is discontinuous there by default.
Key Concepts
Undefined pointsPiecewise definitionFactoring quadratics
Undefined points
When dealing with piecewise functions, understanding where a function is undefined is crucial. An undefined point occurs when you can't calculate a function value due to restrictions like division by zero. In the given exercise, the piecewise function includes a rational expression: \( \frac{x^2 - 3x + 2}{x-2} \). Here, you should always identify values of \( x \) that make the denominator zero since dividing by zero is mathematically impossible. For this function, setting \( x - 2 = 0 \) gives \( x = 2 \), meaning the function will be undefined at this point.
Remember, undefined points are automatically spots of discontinuity in any piecewise function. Hence, ensure to analyze rational expressions for any potential zeroes in the denominator to determine these critical points. This step ensures a thorough understanding of when and why a function "breaks" or becomes discontinuous.
Remember, undefined points are automatically spots of discontinuity in any piecewise function. Hence, ensure to analyze rational expressions for any potential zeroes in the denominator to determine these critical points. This step ensures a thorough understanding of when and why a function "breaks" or becomes discontinuous.
Piecewise definition
Piecewise functions are a type of function defined by different expressions in different parts of their domain. They switch between expressions based on the input value of \( x \). In the given problem, the function \( f(x) \) uses two separate definitions:
Understanding a piecewise definition is key to analyzing function behavior at specific points, like \( x = 1 \) in this case. Here, while approaching \( x = 1 \) from different directions, the values follow \( f(x) = x - 1 \). However, precisely at \( x = 1 \), \( f(x) \) equals 1. This discrepancy between the limit approaching \( x = 1 \) from either side and the actual defined value at \( x = 1 \) leads to a form of discontinuity called a jump discontinuity.
By connecting the behavior of each function piece with the overall continuity, you form a complete picture of the function's nature.
- \( \frac{x^2 - 3x + 2}{x-2} \) for \( x eq 1 \)
- 1 for \( x = 1 \)
Understanding a piecewise definition is key to analyzing function behavior at specific points, like \( x = 1 \) in this case. Here, while approaching \( x = 1 \) from different directions, the values follow \( f(x) = x - 1 \). However, precisely at \( x = 1 \), \( f(x) \) equals 1. This discrepancy between the limit approaching \( x = 1 \) from either side and the actual defined value at \( x = 1 \) leads to a form of discontinuity called a jump discontinuity.
By connecting the behavior of each function piece with the overall continuity, you form a complete picture of the function's nature.
Factoring quadratics
Factoring quadratics is a vital algebraic skill that simplifies expressions, making them easier to analyze. The aim is to express a quadratic equation like \( x^2 - 3x + 2 \) as a product of its linear factors. For this equation, factoring involves finding two numbers that multiply to the constant term (2) and add to the linear coefficient (-3).
Through this method, \( x^2 - 3x + 2 \) is factored into \((x-1)(x-2)\). Applying this to the original function \( \frac{x^2 - 3x + 2}{x-2} \), you can simplify it to \( x - 1 \) as long as you are not dividing by zero at \( x = 2 \). This simplification uncovers any potential undefined points and discontinuities.
Through this method, \( x^2 - 3x + 2 \) is factored into \((x-1)(x-2)\). Applying this to the original function \( \frac{x^2 - 3x + 2}{x-2} \), you can simplify it to \( x - 1 \) as long as you are not dividing by zero at \( x = 2 \). This simplification uncovers any potential undefined points and discontinuities.
- Factoring quadratics is an essential tool, revealing roots and helping explore where expressions may simplify further.
- This approach is not only helpful in solving algebraic equations but also indispensable in analyzing piecewise functions like the one in this problem.
Other exercises in this chapter
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