Problem 58 - Calculus: Graphical, Numerical, Algebraic

Question

Multiple Choice Which of the following statements about the function \(f(x)=\left\\{\begin{array}{ll}{2 x,} & {0 < x < 1} \\ {1,} & {x=1} \\\ {-x+3,} & {1 < x < 2}\end{array}\right.\) is not true? (A) \(f(1)\) does not exist. (B) \(\lim _{x \rightarrow 0^{+}} f(x)\) exists. (C) \(\lim _{x \rightarrow 2^{-}} f(x)\) exists. (D) \(\lim _{x \rightarrow 1} f(x)\) exists. (E) \(\lim _{x \rightarrow 1} f(x)=f(1)\)

Step-by-Step Solution

Answer

The statements A and E are not true about this function.

Step 1: Analyzing statement A

Examine the value for \(f(1)\). It can be observed from the definition of the function that \(f(1) = 1\). Thus, statement A is not true as \(f(1)\) does exist and equals 1.

Step 2: Analyzing statement B

Calculate \(\lim _{x \rightarrow 0^{+}} f(x)\). As x tends to 0 from the positive side, we're under the first condition of the piece-wise function, \(2x\). Hence, the limit is \(2 * 0 = 0\). This means statement B is true

Step 3: Analyzing statement C

Calculate \(\lim _{x \rightarrow 2^{-}} f(x)\). As x tends to 2 from the negative side, we are in the last condition of the piece-wise function, \(-x + 3\). So, the limit is \(-2 + 3 = 1\), which implies that statement C is also true

Step 4: Analyzing statement D

Calculate limit \(\lim _{x \rightarrow 1} f(x)\). To fully analyze, we consider limits from both sides of 1. The limit from the left side (\(x \rightarrow 1^{-}\)) is found under the first condition, \(2x\), giving us \(2 * 1 = 2\). However the limit from the right side (\(x \rightarrow 1^{+}\)) is under the last condition, \(-x + 3\), providing \(-1 +3 = 2\). Both sides provide the same limit, which means that the limit \(\lim _{x \rightarrow 1} f(x)\) does exist. Statement D is true.

Step 5: Analyzing statement E

Check if \(\lim _{x \rightarrow 1} f(x) = f(1)\). From step 1, we know that \(f(1) = 1\). From step 4, we found that \(\lim _{x \rightarrow 1} f(x) = 2\). So, \(\lim _{x \rightarrow 1} f(x) \neq f(1)\) which implies that statement E is not true.

Key Concepts

Limits

Limits are a fundamental concept in calculus, allowing us to understand the behavior of functions as they approach a certain point. In the context of piecewise functions, limits can sometimes be tricky.
It's essential to evaluate limits by considering the function from both sides of a given point. This is especially crucial when the function is split into different expressions, as it is in piecewise scenarios.
For example, for the piecewise function given above, \( f(x)=\left\{ \begin{array}{ll} 2x, & 0 < x < 1 \ 1, & x=1 \ -x+3, & 1 < x < 2 \end{array} \right. \), determining the limit \( \lim_{x \to 1} f(x) \) involves calculating:

• The left-hand limit, \( \lim_{x \to 1^-} f(x) \) using the condition \(2x\), which approaches \(2\).
• The right-hand limit, \( \lim_{x \to 1^+} f(x) \) using the condition \(-x+3\), which also approaches \(2\).

Since both are the same, we conclude that the limit at \(x = 1\) exists and equals \(2\). Understanding these left and right approaching limits helps in verifying function behaviors and is integral when solving piecewise functions.

Continuity

Continuity in a function means there are no jumps, breaks, or holes in the graph of the function around any point in its domain. For a function to be continuous at a point like \(x = 1\), three conditions must be true:

• The function \(f(x)\) is defined at \(x = 1\), meaning \(f(1)\) should exist.
• The limit \(\lim_{x \to 1} f(x)\) must exist.
• The value of the limit must be equal to \(f(1)\).

In the given piecewise function, while we found that both the left-hand and right-hand limits at \(x = 1\) are equal to \(2\), the defined value \(f(1)\) is actually \(1\). This mismatch demonstrates that the function is not continuous at \(x = 1\).
A function's continuity is crucial for understanding how it behaves consistently without unexpected deviations. In piecewise functions, it's common to find discontinuity at the boundaries where the function expression changes.

Calculus

Calculus, as a branch of mathematics, uses limits and continuity to study changes and properties of functions. Within the realm of calculus, piecewise functions are an important study area because they allow us to model situations with different conditions over different intervals.
When dealing with piecewise functions in calculus, we often need to apply different calculus concepts, like limits and continuity, to each piece of the function.
In this example, understanding the different segments of the function \\( f(x)=\left\{ \begin{array}{ll} 2x, & 0 < x < 1 \ 1, & x=1 \ -x+3, & 1 < x < 2 \end{array} \right. \), requires careful differentiation between the sections where each formula applies. Calculus teaches us techniques to:

• Evaluate limits, which help us analyze behavior near boundaries.
• Assess continuity, crucial for ensuring predictions hold true beyond specific intervals.
• Understand the overall behavior of the function, by combining insights from all segments.

Studying these aspects allows for deeper insights into real-world functions that simulate different scenarios through a composite mathematical model.