Problem 4
Question
$$ \begin{aligned} &\text { If } \begin{aligned} f(x) &=\frac{x^{2}+1}{x-1}, \quad x<3 \\ &=\frac{\sin x}{x-3}, \quad x>3, \end{aligned}\\\ &\text { for what values of } x \text { is the function not defined? \\{Ans. } 1,3\\} \end{aligned} $$
Step-by-Step Solution
Verified Answer
The function f(x) is not defined for x = 1 and x = 3. This is because for x < 3, the function is defined as f(x) = \(\frac{x^{2}+1}{x-1}\), which is undefined when x = 1, as the denominator becomes zero. Similarly, for x > 3, f(x) = \(\frac{\sin{x}}{x-3}\), which is undefined when x = 3, again due to the denominator being zero.
1Step 1: Piecewise Function for x<3
For x<3, the function is defined as:
f(x) = \(\frac{x^{2}+1}{x-1}\)
We have a fraction, and we need to determine if there is any value of x that will make the denominator equal to zero, as this would result in an undefined function.
2Step 2: Find Value of x for which Denominator is Zero
Set the denominator (x-1) equal to zero and solve for x:
x-1 = 0
x = 1
So, for x=1, the function is not defined.
3Step 3: Piecewise Function for x>3
For x>3, the function is defined as:
f(x) = \(\frac{\sin{x}}{x-3}\)
Once again, the focus is on the denominator as it can make the function undefined if equal to zero.
4Step 4: Find Value of x for which Denominator is Zero
Set the denominator (x-3) equal to zero and solve for x:
x-3 = 0
x = 3
So, for x=3, the function is not defined in this case as well.
5Step 5: Conclusion
The function f(x) is not defined for the following values of x:
x=1 and x=3
Key Concepts
Understanding Piecewise FunctionsDenominator Equal to Zero and Undefined FunctionsThe Limit of a Function at a Point
Understanding Piecewise Functions
Piecewise functions are mathematical expressions that have different values based on certain conditions. These conditions usually depend on the independent variable, often represented by 'x'. In a piecewise function, we break the whole domain of the function into parts, and each part has its own rule or expression.
For example, in a function like the one provided in the exercise, there are two expressions— one that applies when 'x' is less than 3, and another for when 'x' is greater than 3. This means that the function behaves differently, changing its rule, at certain points. Understanding piecewise functions is crucial for graphing them and analyzing properties like continuity and differentiability.
These types of functions often arise in real-world scenarios. For instance, tax brackets, where the percentage of tax charged depends on where your income falls within certain ranges, are a type of piecewise function. This concept is essential for interpreting various situations that require different outputs given different inputs.
For example, in a function like the one provided in the exercise, there are two expressions— one that applies when 'x' is less than 3, and another for when 'x' is greater than 3. This means that the function behaves differently, changing its rule, at certain points. Understanding piecewise functions is crucial for graphing them and analyzing properties like continuity and differentiability.
These types of functions often arise in real-world scenarios. For instance, tax brackets, where the percentage of tax charged depends on where your income falls within certain ranges, are a type of piecewise function. This concept is essential for interpreting various situations that require different outputs given different inputs.
Denominator Equal to Zero and Undefined Functions
In mathematics, division by zero is undefined because it does not produce a unique finite number. When dealing with functions, especially rational functions like the ones seen in the provided exercise, having a denominator equal to zero means the function is not defined for that particular input value of 'x'.
For instance, with the function \( f(x) = \frac{x^2 + 1}{x - 1} \), setting the denominator \( x - 1 \) to zero and solving for 'x' reveals that \( x = 1 \) yields an undefined expression. Similarly, the function \( f(x) = \frac{\sin{x}}{x - 3} \) becomes undefined when \( x = 3 \) since the denominator becomes zero. This is why ensuring the denominator is never zero is a key aspect of understanding and working with rational functions.
Being alert to the values that create a zero denominator helps prevent miscalculations and misunderstandings of function behavior, which is vital when determining domain and analysing graphs of functions.
For instance, with the function \( f(x) = \frac{x^2 + 1}{x - 1} \), setting the denominator \( x - 1 \) to zero and solving for 'x' reveals that \( x = 1 \) yields an undefined expression. Similarly, the function \( f(x) = \frac{\sin{x}}{x - 3} \) becomes undefined when \( x = 3 \) since the denominator becomes zero. This is why ensuring the denominator is never zero is a key aspect of understanding and working with rational functions.
Being alert to the values that create a zero denominator helps prevent miscalculations and misunderstandings of function behavior, which is vital when determining domain and analysing graphs of functions.
The Limit of a Function at a Point
Although in the given exercise the function is undefined at certain points, the concept of a limit can still give us insight into the behavior of the function as it approaches those points. The limit of a function at a point describes the value that the function approaches as the variable 'x' gets closer and closer to a specific value. It is a fundamental concept in calculus and helps to describe the behavior of functions near points of interest, even if they aren't defined at those points.
In the exercise, while the function is not defined at \( x = 1 \) and \( x = 3 \) due to the zero denominators, we could still explore the limits as \( x \) approaches 1 or 3 from either side. In some cases, this can inform us about potential continuity or the existence of asymptotes.
The limit concept is crucial because it allows mathematicians and scientists to make sensible conclusions about the behavior of functions at points where they cannot be directly evaluated by predicting their behavior based on nearby values.
In the exercise, while the function is not defined at \( x = 1 \) and \( x = 3 \) due to the zero denominators, we could still explore the limits as \( x \) approaches 1 or 3 from either side. In some cases, this can inform us about potential continuity or the existence of asymptotes.
The limit concept is crucial because it allows mathematicians and scientists to make sensible conclusions about the behavior of functions at points where they cannot be directly evaluated by predicting their behavior based on nearby values.
Other exercises in this chapter
Problem 2
Find domain (write conditions only):- i. \(f(x)=\frac{1}{\ln (1-x)}+\sqrt{x+2}\) ii. \(f(x)=\sqrt{\sin ^{-1}\left(\log _{3} x\right)}\) iii. \(\quad f(x)=\sqrt{
View solution Problem 3
$$ \begin{aligned} &\text { Find domain of the function }\\\ &\begin{aligned} f(x) &=\frac{1}{x-3}, \quad x>2 \\ &=\frac{1}{x}, \quad-1
View solution Problem 5
$$ \begin{aligned} &\text { Are the following functions equivalent? }\\\ &\text { i. } \left.\quad f(x)=\frac{x}{x} \text { and } \phi(x)=1 \text { \\{Ans. No }
View solution Problem 6
$$ 3^{-\frac{1}{2} \log _{3} 9} \cdot\left\\{\text { Ans. } \frac{1}{3}\right\\} $$
View solution