AP Calculus AB Exam Question: Consider the function

a) Determine the value of $f(0)$ if it exists, and classify all the points of discontinuity for the function $f(x)$. Justify your answer.

b) On which intervals is the function $f(x)$ continuous? Explain your reasoning.

Answer with Step-by-Step Explanation:

a) To determine the value of $f(0)$, we need to evaluate the function at $x=0$, taking into account the piecewise nature of the function:

For $x<0$, we have $f(x) = 3x+2$, so $f(0) = 3(0)+2 = 2$.

For $x\geq 0$, we have $f(x) = x^2$, but since $0$ is included in this interval, we also consider this piece when evaluating $f(0)$. Therefore, $f(0) = 0^2 = 0$.

Since the two values obtained for $f(0)$ are different (2 and 0), the function $f(x)$ does not have a consistent output at $x=0$. Hence, $f(0)$ does not exist.

To classify the points of discontinuity for the function $f(x)$, we need to analyze the behavior of the function at $x=0$.

For $x<0$, the expression $f(x) = 3x+2$ represents a linear function with a defined slope and a point at $x=0$. Hence, there are no discontinuities in this interval.

For $x\geq 0$, the expression $f(x) = x^2$ represents a quadratic function, which is continuous everywhere. Therefore, there are no discontinuities in this interval.

Overall, the function $f(x)$ is continuous for $x<0$ and $x>0$ (excluding $x=0$). There is a point of discontinuity at $x=0$ because the two pieces of the function have different outputs at this point.

b) The function $f(x)$ is continuous on intervals where it does not have any points of discontinuity. From our analysis in part a), we found that $f(x)$ is continuous for $x<0$ and $x>0$.

Continuity of a function means that the graph can be drawn without lifting the pen or having any breaks, jumps, or holes. In the given function, there are no abrupt changes or jumps in the graph for $x<0$ or $x>0$, and thus, the function is continuous in these intervals.

In conclusion, the function $f(x)$ is continuous for $x<0$ and $x>0$.