Determine the interval(s) on which the following functions are continuous.
f(t)=t+2 / t^2−4

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The function $ f(t) = \frac{t+2}{t^2-4} $ is a rational function, which is continuous everywhere in its domain.

A rational function is undefined where its denominator is zero. Set the denominator equal to zero: $ t^2 - 4 = 0 $.

Factor the equation: $ (t-2)(t+2) = 0 $. This gives the solutions $ t = 2 $ and $ t = -2 $.

The function is undefined at $ t = 2 $ and $ t = -2 $. Therefore, these points are not in the domain of the function.

The function is continuous on the intervals $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Continuity of Functions

A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. For a function to be continuous over an interval, it must be continuous at every point within that interval. This means there are no breaks, jumps, or asymptotes in the function's graph.

Identifying Discontinuities

Discontinuities in a function can occur due to points where the function is undefined, such as division by zero. For the function f(t) = (t + 2) / (t^2 - 4), we need to find values of t that make the denominator zero, as these will indicate points of discontinuity. Factoring the denominator helps identify these critical points.

Intervals of Continuity

Once discontinuities are identified, the next step is to determine the intervals of continuity. This involves analyzing the real number line and excluding the points of discontinuity to find continuous segments. The function is continuous on intervals that do not include these points, which can be expressed in interval notation.

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Related Practice

Textbook Question

Evaluate each limit.

$\lim_{x \to \frac{π}{2}} \frac{\sin (x) - 1}{\sqrt{\sin (x)} - 1}$