Question

Multiple Choice On which of the following intervals is \(f(x)=\frac{1}{\sqrt{x}}\) not continuous? \((\mathbf{A})(0, \infty)\) \((\mathbf{B})[0, \infty)\) \((\mathbf{C})(0,2)\) \((\mathbf{D})(1,2) \quad(\mathbf{E})[1, \infty)\)

Short answer

The function \(f(x)=\frac{1}{\sqrt{x}}\) is not continuous on the interval (\mathbf{B})[0, \infty).

Step-by-step solution

Analyzing the Options

Out of the five provided options, look for the interval that includes 0. That will be an interval where the function is not continuous since \(\frac{1}{\sqrt{x}}\) is undefined at x = 0.

Identifying the Correct Interval

From the provided options, only one interval, ((\mathbf{B})[0, \infty), includes 0. In this interval, x starts from 0 (which includes 0). For all values of \(x \geq 0\), \(f(x) =\frac{1}{\sqrt{x}}\) is undefined at 0 making the function discontinuous.

Choosing the Correct Answer

The interval that contains 0 and thus represents where f(x) is discontinuous is option (\mathbf{B})[0, \infty).

Key concepts

Continuous Functions

Understanding the behavior of continuous functions is a fundamental aspect of analyzing mathematical relationships. A function is said to be continuous at a point if the following three conditions are met:

• The function is defined at the point.
• The limit of the function as it approaches the point exists.
• The limit of the function is equal to the function’s value at that point.

In simpler terms, you can think of a continuous function as a smooth curve with no breaks, jumps, or holes at any point within its domain. For the function in question, \( f(x)=\frac{1}{\sqrt{x}} \), it is continuous where it is defined—that is, for all positive values of x.However, for continuous functions dealing with square roots, like our example, an important thing to note is that the domain cannot include values that would result in taking the square root of a negative number, as this would be undefined within the real number system. This is why we systematically sift through options to exclude any intervals that include non-permissible values for x.

Discontinuities in Functions

When we talk about discontinuities in functions, we are referring to points where a function behaves erratically or unpredictably. Discontinuities are essentially the opposites of what makes a function continuous. There are several types of discontinuities, the most common being:

• Point discontinuity: The function is defined at a point but is not continuous there.
• Jump discontinuity: The function has a sudden jump in values at a point.
• Infinite discontinuity: The function approaches infinity at a point.

For the function \( f(x)=\frac{1}{\sqrt{x}} \), we have an example of an infinite discontinuity at x=0 since as x approaches 0, the values of the function increase without bound. Thus, in the interval option \( [0, \infty) \), the inclusion of x=0 leads to the function being not only undefined but also discontinuous at that point due to the infinite nature of the function value.

Limits and Continuity

The concept of limits is intimately connected with understanding continuity. The limit of a function describes the behavior of the function as the input approaches a particular value. In continuity, a function at a point is determined by evaluating the limit of the function as it approaches that point from both sides.When a function's limit matches the actual value of the function at a point, the function is continuous at that point. If the limit does not exist or is different from the function's actual value, we are faced with a discontinuity.To apply this to our example, we would find the limit of \( f(x)\) as x approaches any given point within its domain. Everywhere except at x=0, the limits exist and match the function's value, confirming continuity. At x=0, the limit of \( f(x)\) as it approaches from the positive side is infinity, which means that \( f(x)\) is not continuous at x=0. Therefore, when assessing functions for continuity, one must closely consider their limits around points of interest—especially where the function's formula might lead to undefined or infinite results.