Problem 5 - The Calculus with Analytic Geometry

Question

Determine whether the function is continuous or discontinuous on each of the indicated intervals. $$ f(t)=\frac{|t-1|}{t-1} ;(-\infty, 1),(-\infty, 1],[-1,1],(-1,+\infty),(1,+\infty) $$

Step-by-Step Solution

Answer

Continuous on \((-\infty, 1)\) and \((1, +\infty)\). Discontinuous on \((-\infty, 1]\), \([-1, 1]\), and \((-1, +\infty)\).

Step 1 - Analyze Function Behavior

The function given is \( f(t) = \frac{|t-1|}{t-1} \). To understand its behavior, let's simplify it in terms of intervals depending on the value of \( t \) relative to 1.

Step 2 - Consider Interval \( t < 1 \)

When \( t < 1 \), \( |t-1| = -(t-1) \). Thus, the function in this interval becomes \( f(t) = \frac{-(t-1)}{t-1} = -1 \). This shows that \( f(t) = -1 \) for all \( t < 1 \). This function is continuous in \( (-\infty, 1) \) since it is a constant function.

Step 3 - Consider Interval \( t > 1 \)

When \( t > 1 \), \( |t-1| = t-1 \), so \( f(t) = \frac{t-1}{t-1} = 1 \). This implies \( f(t) = 1 \) for all \( t > 1 \). This function is again continuous in \( (1, \infty) \) since it is a constant.

Step 4 - Consider \( t = 1 \)

At \( t = 1 \), the function \( f(t) \) is not defined as it leads to division by zero ( \( \frac{|t-1|}{t-1} = \frac{0}{0} \)). Therefore, the function is discontinuous at \( t = 1 \).

Step 5 - Analyze Interval \( (-\infty, 1] \)

This interval includes \( t = 1 \) where the function is not defined. Hence, \( f(t) \) is discontinuous in \( (-\infty, 1] \).

Step 6 - Analyze Interval \( [-1, 1] \)

Similarly, this interval includes \( t = 1 \) where the function is not defined. Thus, \( f(t) \) is discontinuous in \( [-1, 1] \).

Step 7 - Analyze Interval \( (-1, +\infty) \)

As \( t > 1 \) is included and there is no discontinuity for any value less than \( t = 1 \), the function is discontinuous in \( (-1, +\infty) \).

Step 8 - Analyze Interval \( (1, +\infty) \)

Since the interval starts right after \( t = 1 \), and all points in \( (1, \infty) \) have a continuous value of 1, the function is continuous in \( (1, +\infty) \).

Key Concepts

discontinuous functions

A function is considered discontinuous at a point if there is a sudden jump or there exists a value where the function is not defined. For our function, \( f(t) = \frac{|t-1|}{t-1} \), we observe that it behaves differently at \( t = 1 \). This point creates a disruption in the function because \( f(1) \) leads to \( \frac{0}{0} \), which is undefined. Therefore, \( f(t) \) is discontinuous at \( t = 1 \). This discontinuity might affect various intervals that include this point.

interval analysis

When examining the continuity of a function over an interval, we analyze the behavior and definition of the function across that range. For instance, evaluating \( f(t) = \frac{|t-1|}{t-1} \) over different intervals:

• For \( t < 1 \), the function simplifies to \( f(t) = -1 \), which is constant and thus continuous in \( (-\infty, 1) \).
• For \( t > 1 \), the function changes to \( f(t) = 1 \), again constant and continuous in \( (1, \infty) \).
• When the interval includes \( t = 1 \) such as \( (-\infty, 1] \) or \( [-1, 1] \), the function becomes discontinuous due to the undefined nature at \( t = 1 \).
• Intervals that do not include 1 but cross it, such as \( (-1, +\infty) \), also consider the discontinuity at 1 making the overall analysis critical to understanding continuity or discontinuity in specified ranges.

function behavior

Understanding a function's behavior helps in determining its continuity or discontinuity. Our function, \( f(t) = \frac{|t-1|}{t-1} \), shows different values based on the domain of \( t \). This change portrays how absolute value and division impact the function's output:

• For \( t < 1 \), the function consistently outputs -1, showing constant behavior.
• For \( t > 1 \), the function shifts to a constant output of 1.
• At \( t = 1 \), both conditions merge causing an undefined value (\( \frac{0}{0} \)), resulting in a discontinuous behavior.

By breaking the function into respective intervals and understanding its behavior, we identify continuous segments (constant outputs) from discontinuous points (undefined values).