Problem 57 - Calculus: Graphical, Numerical, Algebraic

Question

Multiple Choice Which of the following points is not a point of discontinuity of \(f(x)=\sqrt{x-1} ?\) (A) \(x=-1 \quad\) (B) \(x=-1 / 2 \quad\) (C) \(x=0\) (D) \(x=1 / 2 \quad\) (E) \(x=1\)

Step-by-Step Solution

Answer

The point \(x=1\) (Option E) is not a point of discontinuity for the function \(f(x)=\sqrt{x-1}\).

Step 1: Identify the solved domain of the function

The function \(f(x)=\sqrt{x-1}\) is only defined for \(x \geq 1\). This means the function is continuous for \(x \geq 1\). Therefore, any value that is less than 1, is a point of discontinuity for this function.

Step 2: Evaluate the given options for the definition of the function

Among the given options, all values of x are less than 1, except for option E that is \(x=1\). Since \(x=1\) falls in the solved domain of the function \(\sqrt{x-1}\), this value does not represent a discontinuity of the function.

Key Concepts

Domain of a function

The domain of a function is an essential mathematical concept to understand. It represents all the possible input values that a function can accept without causing any mathematical errors. For instance, if we have a function involving division, we must exclude values that would make the denominator zero. Similarly, for functions involving square roots, the values under the root must be non-negative because square root operations on negative numbers result in complex numbers.
Let's look at the function in our exercise: \(f(x) = \sqrt{x-1}\). To find its domain, determine where the expression inside the square root is non-negative. So, we solve \(x-1 \geq 0\), which simplifies to \(x \geq 1\). This tells us that the function is only defined for values of \(x\) that are equal to or greater than 1.
Understanding the domain helps us know the behavior of the function and identify any input values that would lead to undefined points or discontinuities.

Discontinuity

In mathematics, discontinuity refers to points on the graph of a function where the function is not continuous. A function might suddenly jump, break, or take a sharp turn at these points. Finding and understanding discontinuities is crucial in calculus and real analysis.It’s essential to check the domain first, as any value not included in the domain will automatically be a point of discontinuity. For our specific function, \(f(x) = \sqrt{x-1}\), the domain is \(x \geq 1\). This means any \(x\) less than 1 is not defined, thus, a discontinuity occurs at these points.
In our exercise, every \(x\) value in options A to D is less than 1. These are discontinuities. Only the value \(x = 1\) (option E) is defined within the function's domain. Recognizing discontinuities helps us plot functions accurately and understand their behavior better.

Square root function

A square root function is a type of radical function represented as \(f(x) = \sqrt{x}\). It’s a versatile and essential function encountered frequently in mathematics. Its graph typically resembles a curve that starts at the origin (or a certain point) and increases. However, it's crucial to note that square root functions are not defined for negative values. This is because the square root of a negative number is not a real number, thus limiting the domain.For \(f(x) = \sqrt{x-1}\), the graph shifts one unit to the right, meaning the starting point is at \(x = 1\). Understanding square root functions involves:

• Recognizing the transformation from the basic form \(\sqrt{x}\) to \(\sqrt{x-1}\).
• Realizing how the domain changes due to the transformation (ensuring that what’s under the root is non-negative).
• Knowing that these functions continuously increase as \(x\) increases.

Grasping these concepts allows you to scrutinize the function's graph, identify its fundamental properties, and solve any potential problems with ease.