Question

Discuss the continuity of the function on the closed interval. $$ \begin{array}{lll} \text { Function } & & \text { Interval } \\ f(t)=2-\sqrt{9-t^{2}} & & {[-2,2]} \end{array} $$

Short answer

The function \(f(t)=2-\sqrt{9-t^{2}}\) is continuous on the interval [-2,2].

Step-by-step solution

- Define the domain of the function

The equation inside the square root, \(9-t^{2}\), must be greater than or equal to 0 (since the square root of a negative number is undefined). This gives a possible function domain of \(-3 \leq t \leq 3\).

- Identify the relevant interval for the function

The given interval is [-2,2]. The function will only be evaluated within this interval, even though the function's natural domain extends beyond this. The exercise asks for continuity over the specific interval [-2,2].

- Analyze the continuity over the given interval

According to the definition of continuity, a function is said to be continuous on an interval if it is continuous at every point in that interval. Comparing the possible domain from Step 1 with the given interval, every value in the interval [-2,2] is also in the interval [-3,3] which is the function's domain. Therefore, at every point in [-2,2], the function \(f(t)=2-\sqrt{9-t^{2}}\) is defined and continuous.