Give an example of a function f which satisfies the initial value problemI.V.P on a closed and bounded interval [a, b], but is not continuous on [a, b]

One classic example of a function $f$ that satisfies an initial value problem (IVP) on a closed and bounded interval $[a, b]$, but is not continuous on $[a, b]$ is given by the following:

Example Function

Let $f(x) = \begin{cases} 1 & \text{if } x \in [a, c) \\ 0 & \text{if } x = c \\ 1 & \text{if } x \in (c, b] \end{cases}$ where $a < c < b$.

Initial Value Problem

Consider the IVP defined by:

1. Differential Equation: $f'(x) = 0$ for all $x \neq c$
2. Initial Condition: $f(a) = 1$

Analysis of the Function

1. Piecewise Definition: The function is piecewise constant: it takes the value 1 on the intervals $[a, c)$ and $(c, b]$, but is defined to be 0 at the point $x = c$.
2. Discontinuity: The function is discontinuous at the point $x = c$, where the function value jumps from 1 to 0.
3. Closed and Bounded Interval: The interval $[a, b]$ is closed and bounded as per the definition.

This function meets the requirements of the IVP at all points except for $c$, where it is not continuous. The derivatives exist wherever it is defined, leading to the conclusion that the function satisfies the initial value problem despite the discontinuity.

Conclusion

Thus, $f(x)$ satisfies the IVP condition on $[a, b]$ but is not continuous over the interval due to the jump discontinuity at $c$.