Problem 61
Question
Is the function given by $$ G(x)=\left\\{\begin{array}{ll} \frac{x^{2}-3 x-4}{x-4}, & \text { for } x<4 \\ 2 x-3, & \text { for } x \geq 4 \end{array}\right. $$ continuous at \(x=4 ?\) Why or why not?
Step-by-Step Solution
Verified Answer
Yes, the function \( G(x) \) is continuous at \( x = 4 \) because \( G(4) = \lim_{x \to 4^-} G(x) = \lim_{x \to 4^+} G(x) = 5 \).
1Step 1: Understand the Problem
We are given a piecewise function \( G(x) \) and need to determine if it is continuous at \( x = 4 \). A function is continuous at a point if the following three conditions are met: (1) \( G(4) \) is defined, (2) \( \lim_{x \to 4^-} G(x) \) exists and equals \( G(4) \), and (3) \( \lim_{x \to 4^+} G(x) \) equals \( G(4) \).
2Step 2: Evaluate \( G(4) \)
Examine the piecewise condition for \( G(x) \) when \( x \geq 4 \). The relevant formula is \( 2x-3 \). Thus, \( G(4) = 2(4)-3 = 8-3 = 5 \). So, \( G(4) \) is 5.
3Step 3: Calculate \( \lim_{x \to 4^-} G(x) \)
Use the formula for \( x < 4 \), which is \( \frac{x^2-3x-4}{x-4} \), to find the limit as \( x \) approaches 4 from the left. Factor the numerator: \( x^2-3x-4 = (x-4)(x+1) \). So, \( G(x) = x+1 \) when \( x eq 4 \). The limit is \( \lim_{x \to 4^-} (x+1) = 5 \).
4Step 4: Calculate \( \lim_{x \to 4^+} G(x) \)
For \( x \to 4^+ \), we use the formula \( 2x-3 \). The limit is \( \lim_{x \to 4^+} (2x-3) = 2(4)-3 = 5 \).
5Step 5: Determine Continuity at \( x = 4 \)
Check if all conditions for continuity are met: \( G(4) = 5 \), \( \lim_{x \to 4^-} G(x) = 5 \), and \( \lim_{x \to 4^+} G(x) = 5 \). All three values are equal, so \( G(x) \) is continuous at \( x = 4 \).
Key Concepts
Piecewise FunctionLimitsContinuity Conditions
Piecewise Function
A piecewise function is a type of function that is defined by different expressions depending on the region of the input. Picture it as a piece of a puzzle; each piece or segment is a separate function with its individual rule. However, when combined, they create the larger picture, which in this case is the full range of the function.
For example, the function \( G(x) \) in our problem is a piecewise function. For inputs less than 4, it is defined by \( \frac{x^{2}-3x-4}{x-4} \), and for inputs equal to or greater than 4, it shifts to the rule \( 2x-3 \). This characteristic of changing rules based on input values is what makes the function piecewise.
When analyzing a piecewise function, it's crucial to understand how these different segments interact, especially around transition points, such as at \( x=4 \) in our example.
For example, the function \( G(x) \) in our problem is a piecewise function. For inputs less than 4, it is defined by \( \frac{x^{2}-3x-4}{x-4} \), and for inputs equal to or greater than 4, it shifts to the rule \( 2x-3 \). This characteristic of changing rules based on input values is what makes the function piecewise.
When analyzing a piecewise function, it's crucial to understand how these different segments interact, especially around transition points, such as at \( x=4 \) in our example.
Limits
Limits are a fundamental concept in calculus, used to determine the value that a function approaches as the input approaches a certain point. This is particularly useful for piecewise functions at their boundaries.
In our exercise, we investigated the limits approaching the point \( x=4 \) from both the left (\( 4^- \)) and the right (\( 4^+ \)).
- To find the left-hand limit (\( \lim_{x \to 4^-} G(x) \)), we looked at the function defined for \( x < 4 \). By simplifying \( \frac{x^{2}-3x-4}{x-4} \), we found \( x+1 \) and concluded that as \( x \to 4^- \), the limit is 5.
- For the right-hand limit (\( \lim_{x \to 4^+} G(x) \)), we used the function \( 2x-3 \) since it applies for \( x \geq 4 \) and found that as \( x \to 4^+ \), the limit is also 5.
Both limits needed to be equal for there to be a possibility of continuity at \( x=4 \). They show how smoothly functions transition between different sections or rules, which is key in analyzing piecewise functions.
In our exercise, we investigated the limits approaching the point \( x=4 \) from both the left (\( 4^- \)) and the right (\( 4^+ \)).
- To find the left-hand limit (\( \lim_{x \to 4^-} G(x) \)), we looked at the function defined for \( x < 4 \). By simplifying \( \frac{x^{2}-3x-4}{x-4} \), we found \( x+1 \) and concluded that as \( x \to 4^- \), the limit is 5.
- For the right-hand limit (\( \lim_{x \to 4^+} G(x) \)), we used the function \( 2x-3 \) since it applies for \( x \geq 4 \) and found that as \( x \to 4^+ \), the limit is also 5.
Both limits needed to be equal for there to be a possibility of continuity at \( x=4 \). They show how smoothly functions transition between different sections or rules, which is key in analyzing piecewise functions.
Continuity Conditions
A function is continuous at a point if its behavior at that point matches its behavior as inputs approach from both directions. In more technical terms, continuity at a particular point involves three key conditions.
1. The function has a defined value at the point: In our example, \( G(4) \) is defined since \( 2(4)-3 = 5 \). So, \( G(4) = 5 \).
2. The left-hand limit equals the function's value at the point: As determined earlier, \( \lim_{x \to 4^-} G(x) = 5 \).
3. The right-hand limit also equals the function's value at the point: We found \( \lim_{x \to 4^+} G(x) = 5 \).
Since all three conditions are met, \( G(x) \) is continuous at \( x=4 \). This step-by-step verification of each condition demonstrates a systematic way of establishing continuity, which is particularly essential when dealing with piecewise functions. Continuity ensures that there are no holes or jumps at the point of interest, ensuring a smooth graph across that region.
1. The function has a defined value at the point: In our example, \( G(4) \) is defined since \( 2(4)-3 = 5 \). So, \( G(4) = 5 \).
2. The left-hand limit equals the function's value at the point: As determined earlier, \( \lim_{x \to 4^-} G(x) = 5 \).
3. The right-hand limit also equals the function's value at the point: We found \( \lim_{x \to 4^+} G(x) = 5 \).
Since all three conditions are met, \( G(x) \) is continuous at \( x=4 \). This step-by-step verification of each condition demonstrates a systematic way of establishing continuity, which is particularly essential when dealing with piecewise functions. Continuity ensures that there are no holes or jumps at the point of interest, ensuring a smooth graph across that region.
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