Determine the values of x, if any, at which each function is discontinuous. At each number where f is discontinuous, state the condition(s) for continuity that are violated.

f(x)= x+5 if x is less than or =0
-x^2+5 if x>0

Who says it is discontinuous ? Looks continuous to me.

How did you find that it was continuous?

To determine the values of x at which the function f(x) is discontinuous, we have to examine the conditions for continuity.

Condition 1: The function f(x) is continuous if the limits from the left and right side of a point x are equal to each other.
Condition 2: The function f(x) is continuous if the value of f(x) at x is equal to the limit of f(x) as x approaches that point.

Now let's analyze the function f(x) step by step:

For x ≤ 0:
Since f(x) = x + 5 when x is less than or equal to 0, there are no conditions violated, and the function is continuous for all values of x less than or equal to 0.

For x > 0:
We have f(x) = -x^2 + 5 when x is greater than 0.
To check continuity, we consider the conditions:

Condition 1: The limit of f(x) as x approaches 0 from the left side:
lim(x→0-) (-x^2 + 5) = -0^2 + 5 = 5

Condition 2: The limit of f(x) as x approaches 0 from the right side:
lim(x→0+) (-x^2 + 5) = -0^2 + 5 = 5

Condition 1 is satisfied since the limits are equal on both sides of x = 0.

Condition 2: The value of f(x) at x = 0:
f(0) = 0 + 5 = 5

Here, Condition 2 is also satisfied since the value of f(x) at x = 0 is equal to the limits.

Hence, the function f(x) is continuous for all x, including 0.

Therefore, the function f(x) is continuous for all values of x, and there are no values of x at which it is discontinuous.

To determine the values of x at which the function f(x) is discontinuous, we need to look for any points where the function is defined differently or where there are jumps or holes in the graph.

For the given function, f(x) is defined in two separate cases:

1. When x ≤ 0: In this case, f(x) = x + 5. This is a linear function, and for any value of x, the function is continuous. There are no discontinuities in this part of the function.

2. When x > 0: In this case, f(x) = -x^2 + 5. This is a quadratic function. Quadratic functions are always continuous, so there are no general discontinuities for this part of the function. However, there might be specific points where the equation changes.

Since the function is continuous for each specific expression, we only need to check if the function is defined differently at specific points for each condition.

Now, let's consider when the transition from x ≤ 0 to x > 0 occurs.

At the point x = 0, there is a transition between the two cases. When x = 0, the function is defined as f(0) = 0 + 5 which is equal to 5. This means that at x = 0, the value of f(x) changes from x + 5 (for x ≤ 0) to -x^2 + 5 (for x > 0).

Since there is a change in the formula at x = 0, we can say that f(x) is discontinuous at x = 0. The condition for continuity that is violated is that the function should have a unique value at x = 0, but in this case, it has two different expressions depending on the side of x.

To summarize:
- The function f(x) is continuous for all values of x except at x = 0.
- At x = 0, f(x) is discontinuous, violating the condition of having a unique value.

Therefore, the function f(x) has a discontinuity at x = 0.