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 \(0 \leq t \leq 3\).