Question

Discuss the continuity of the function on the closed interval. $$ \begin{array}{lll} \text { Function } & \text { Interval } \\ g(x)=\frac{1}{x^{2}-4} &{[-1,2]} \end{array} $$

Short answer

The function \(g(x) = \frac{1}{x^{2}-4}\) is discontinuous on the interval \([-1,2]\) due to a discontinuity at \(x = 2\).

Step-by-step solution

Checking the denominator

First, check to see whether the denominator \(x^{2}-4\) equals zero in the interval. It is equal to zero when \(x = 2\) or \(x = -2\). Since -2 is not in the interval \([-1,2]\), and 2 is a limit point, the function will be considered on \([-1,2)\

Calculating the limit

Calculate the limit as \(x\) approaches 2 from the left side. The limit is \(\lim_{x \to 2^{-}} g(x)=\lim_{x \to 2^{-}} \frac{1}{x^{2}-4}\) which evaluates to -∞. It implies that the function goes to negative infinity as \(x\) approaches 2 from the left.

Finding the function value at 2

Next, observe that the function is undefined when \(x=2\), as this makes the denominator zero and produces an undefined expression.

Discussion of Continuity

Since the limit as \(x\) approaches 2 from the left is not equal to the function value at \(x=2\) (which doesn't exist), the function is not continuous at \(x=2\). Therefore, the function \(g(x) = \frac{1}{x^{2}-4}\) is discontinuous on the interval \([-1,2]\).

Key concepts

Limits and Continuity

Understanding the continuity of a function starts by looking at limits. A function is continuous at a point if the limit of the function as it approaches that point from both sides equals the function's value at that point. In simpler terms, a continuous function has no breaks, jumps, or holes at that specific point.

When dealing with a function like \(g(x) = \frac{1}{x^{2} - 4}\), assessing limits is crucial because it tells us what happens to the function as we get close to certain points - such as \(x = 2\) in this case. The fact that the limit as \(x\) approaches 2 from the left goes to negative infinity (\(-\infty\)) highlights a break in continuity; the function does not approach a real number, indicating a vertical asymptote and thus a point of discontinuity.

Moreover, for a function defined on a closed interval, we expect it to be continuous through the entire interval for it to be deemed wholly continuous. If there’s a single point where the function doesn’t meet the continuity requirements, as we found at \(x = 2\) with our function \(g(x)\), then the function is not continuous on the interval.

Closed Interval Analysis

Closed interval analysis involves studying the behavior of a function within a specific range where both endpoints are included. This is important because the behavior of functions can differ greatly at the boundaries of an interval. To say a function is continuous on a closed interval \([a, b]\), it must be continuous at every point inside the interval and it should have finite one-sided limits at the endpoints.

In the case of the function \(g(x)\), while we look at the closed interval \([-1, 2]\), we must carefully handle the endpoint \(x = 2\). Although \(2\) is included in the notation of the interval, the function itself is undefined at \(2\) due to a zero in the denominator. This single point disrupts the continuity of the function over the entire interval because the condition for continuity is not fulfilled at this endpoint. Thus, even if a function behaves well throughout the rest of the interval, just one problematic point – especially at an endpoint – can alter our conclusion regarding its continuity.

Discontinuous Functions

Discontinuous functions exhibit breaks, jumps, or holes in their graph. These disruptions may occur due to an undefined point, such as a division by zero or when limits do not match up with the function's value at a certain point.

A classic example of a discontinuous function at a point is presented by \(g(x) = \frac{1}{x^{2} - 4}\) at \(x = 2\), which is an endpoint of the interval \([-1, 2]\). At this point, the denominator becomes zero, thus making the function value undefined, and highlighting a type of discontinuity known as an infinite discontinuity because the limit as \(x\) approaches the point is infinite.

Distinguishing between various types of discontinuities helps us understand how a function behaves and allows for better predictions of its behavior nearby the points of discontinuity. In many mathematical contexts, identifying these points and understanding their nature is key for further analyses such as integration and curve sketching.