Question

Sketch the graph of the given function. In each case determine whether \(f\) is continuous, piecewise continuous, or neither on the interval \(0 \leq t \leq 3 .\) $$ f(t)=\left\\{\begin{array}{ll}{t^{2},} & {0 \leq t \leq 1} \\ {2+t,} & {1

Short answer

Given a function defined by: - \(f(t) = t^2\) for \(0 \leq t \leq 1\) - \(f(t) = 2+t\) for \(1 < t \leq 2\) - \(f(t) = 6-t\) for \(2 < t \leq 3\) After analyzing and sketching the function, we found that the function is discontinuous at \(t=1\) but continuous at \(t=2\). Therefore, the function is considered piecewise continuous on the interval \(0 \leq t \leq 3\).

Step-by-step solution

Sketch the graph of each piece of the function

We need to sketch the graph of the function in each of the given intervals: \(t^2\) for \(0 \leq t \leq 1\), \(2+t\) for \(1<t\leq 2\), and \(6-t\) for \(2<t\leq 3\).

Determine continuity at transition points

Now, we need to check the continuousness of the function at the transition points between the intervals. There are two points to check - \(t=1\) and \(t=2\). For \(t=1\), we have: - From the first interval: \(f(t) = t^2 \Rightarrow f(1) = 1^2 = 1\) - From the second interval: \(f(t) = 2 + t \Rightarrow f(1) = 2 + 1 = 3\) The function is discontinuous at \(t=1\) because the limits from the left and right do not match (\(1 \neq 3\)). For \(t=2\), we have: - From the second interval: \(f(t) = 2 + t \Rightarrow f(2) = 2 + 2 = 4\) - From the third interval: \(f(t) = 6 - t \Rightarrow f(2) = 6 - 2 = 4\) The function is continuous at \(t=2\) because the limits from the left and right match (\(4 = 4\)).

Determine the classification of the function on the interval \(0 \leq t \leq 3\)

Since the function is discontinuous at \(t=1\), it is not continuous on the interval \(0 \leq t \leq 3\). However, the function has only one point of discontinuity, so it is piecewise continuous on the interval \(0 \leq t \leq 3\).

Key concepts

Graph Sketching

Graph sketching is an essential first step when analyzing piecewise continuous functions. It helps us visualize how different segments of a function behave within given intervals. In our case, the function \(f(t)\) is defined as three separate expressions in the intervals \(0 \leq t \leq 1\), \(1 < t \leq 2\), and \(2 < t \leq 3\). Each interval corresponds to a unique mathematical expression:

• \(t^2\) for \(0 \leq t \leq 1\): This is a parabola segment opening upwards.
• \(2+t\) for \(1 < t \leq 2\): This is a linear line segment with a slope of 1.
• \(6-t\) for \(2 < t \leq 3\): This is another linear segment but has a slope of -1.

To effectively sketch this graph:- Start by plotting each segment accurately within its interval.- Ensure the endpoints of each interval are correctly plotted.- Carefully illustrate the changes or breaks at transition points where the expression changes, as these are key in understanding the function's behavior.

Function Continuity

Continuity in a function implies there is no abrupt change in its value within its domain. When analyzing each part of our piecewise function, we focus on whether the transition is smooth between intervals.
In simple terms, a function is continuous at a point if:

• The limit from the left of the point equals the limit from the right.
• This common limit equals the function's value at the point.

In our function, we identify points \(t=1\) and \(t=2\) as transition zones where each segment of the function might introduce discontinuity. At \(t=1\), the two different segment results \(1^2 = 1\) versus \(2+1 = 3\) indicate a jump, hence it's not continuous. However, at \(t=2\), both segments smoothly connect giving the same function value, so no continuity issue arises here. Understanding these details is crucial for determining how the entire function behaves across \(0 \leq t \leq 3\).

Transition Points Analysis

Transition points are key locations where changes in functional expressions occur, often leading to changes in the function's behavior. In our scenario, transition points are at \(t=1\) and \(t=2\).
Analyzing these points involves:

• Evaluating the limit from the left and right side of the transition point.
• Checking if these limits, and the value at the transition point, align or differ.

For \(t=1\), the limits on either side don't match (left = 1, right = 3), signaling a discontinuity. For \(t=2\), both limits reach 4 consistently, maintaining continuity. Identifying and understanding these transition points is vital, as they can dramatically affect whether a function is deemed piecewise continuous or discontinuous overall. Transition points thus form the backbone of continuity analysis in piecewise functions.

Interval Analysis

Interval analysis breaks down the entire domain of a function into smaller parts to study its behavior more deeply, which is incredibly useful for piecewise functions.
In this context, we're interested in the intervals \(0 \leq t \leq 1\), \(1 < t \leq 2\), and \(2 < t \leq 3\). For each segment, the focus is:

• Evaluating the individual function piece on its respective interval.
• Observing how these segments gel together at their boundaries.

During our analysis, we note the unique expression for each interval and judge how they contribute to the entire graph's continuity:- Each section operates continuously within itself.- It is at their boundaries—transition points—where the overall function's behavior is observed.Interval analysis ensures one grasps the detailed and expansive picture of how seamless or broken a function might be, helping to label it correctly as either continuous, piecewise continuous, or neither.