Determine if Continuous f(x)=(x^2+1)/(x-1)
The question is asking whether the function f(x) = (x^2 + 1) / (x - 1) is continuous. Continuity of a function typically means that the function does not have any breaks, jumps, or holes at any point in its domain. For this specific function, you would need to investigate if it is continuous for all values of x, and if not, identify the points where the function is not continuous and explain why. In the context of this function, special attention would likely be required at x = 1, since that is where the denominator of the fraction becomes zero, which might cause a discontinuity.
$f \left(\right. x \left.\right) = \frac{x^{2} + 1}{x - 1}$
Answer
Solution:
Step:1
Identify the domain where the function is defined to check its continuity.
Step:1.1
To find the values that make the function undefined, equate the denominator of $\frac{x^{2} + 1}{x - 1}$ to zero: $x - 1 = 0$.
Step:1.2
Solve the equation by adding $1$ to both sides: $x = 1$.
Step:1.3
The domain consists of all $x$ values that do not make the denominator zero.
In Interval Notation: $(-\infty, 1) \cup (1, \infty)$ In Set-Builder Notation: $\{ x | x \neq 1 \}$
Step:2
Given that the domain excludes $x = 1$, the function $\frac{x^{2} + 1}{x - 1}$ is not continuous for all real numbers.
Step:3
Knowledge Notes:
To determine if a function $f(x)$ is continuous, we need to ensure that it is defined for all values within its domain, and there are no breaks, jumps, or holes in its graph. A function is continuous at a point $x = a$ if the following conditions are met:
$f(a)$ is defined.
The limit of $f(x)$ as $x$ approaches $a$ exists.
The limit of $f(x)$ as $x$ approaches $a$ is equal to $f(a)$.
For rational functions like $f(x) = \frac{x^2 + 1}{x - 1}$, discontinuities can occur where the denominator is zero since division by zero is undefined. Therefore, finding the domain of the function is a crucial step in determining its continuity.
In this case, the denominator $x - 1$ is zero when $x = 1$. Thus, the function is not defined at $x = 1$, and there is a discontinuity at this point. The domain of the function is all real numbers except $x = 1$. This can be expressed in interval notation as $(-\infty, 1) \cup (1, \infty)$, which means all real numbers less than 1 and all real numbers greater than 1. In set-builder notation, it is written as $\{ x | x \neq 1 \}$, which reads as "the set of all $x$ such that $x$ is not equal to 1."
Since the function is not defined at $x = 1$, it is not continuous over the entire set of real numbers. The function is continuous on its domain, but because its domain does not include all real numbers, the function itself is not continuous over the real numbers.
