Identifying Potential Discontinuities

Oct 13, 2025 by orovajewels.com 40 views

Determine the Values of $x$ for Which the Function $f(x) = \frac{6}{\sqrt{x+13}}$ is Continuous

In mathematics, a function is considered continuous if it can be drawn without lifting the pencil from the paper. In other words, a function is continuous if its graph has no gaps or jumps. In this article, we will determine the values of $x$ for which the function $f(x) = \frac{6}{\sqrt{x+13}}$ is continuous.

A function $f(x)$ is said to be continuous at a point $x=a$ if the following conditions are satisfied:

The function is defined at $x=a$ , i.e., $f(a)$ is defined.
The limit of the function as $x$ approaches $a$ exists, i.e., $\lim_{x \to a} f(x)$ exists.
The limit of the function as $x$ approaches $a$ is equal to the value of the function at $x=a$ , i.e., $\lim_{x \to a} f(x) = f(a)$ .

The given function is $f(x) = \frac{6}{\sqrt{x+13}}$ . To determine the values of $x$ for which this function is continuous, we need to analyze the function and identify any points where it may not be continuous.

Identifying Potential Discontinuities

The function $f(x) = \frac{6}{\sqrt{x+13}}$ has a square root in the denominator. The square root of a negative number is not defined in the real number system, so we need to be careful when dealing with negative values of $x+13$ . Additionally, the function is not defined when the denominator is equal to zero, i.e., when $x+13=0$ .

Solving for Potential Discontinuities

To find the potential discontinuities, we need to solve the equation $x+13=0$ . This equation has a solution of $x=-13$ . Therefore, the function $f(x) = \frac{6}{\sqrt{x+13}}$ is not defined at $x=-13$ .

Now that we have identified the potential discontinuity at $x=-13$ , we need to determine the values of $x$ for which the function is continuous. To do this, we need to analyze the behavior of the function as $x$ approaches $-13$ .

Left-Hand Limit

To find the left-hand limit of the function as $x$ approaches $-13$ , we can use the following limit:

\lim_{x \to -13^-} f(x) = \lim_{x \to -13^-} \frac{6}{\sqrt{x+13}}

Since the square root of a negative number is not defined in the real number system, the left-hand limit does not exist.

Right-Hand Limit

To find the right-hand limit of the function as $x$ approaches $-13$ , we can use the following limit:

\lim_{x \to -13^+} f(x) = \lim_{x \to -13^+} \frac{6}{\sqrt{x+13}}

Since the square root of a negative number is not defined in the real number system, the right-hand limit does not exist.

Based on our analysis, we can conclude that the function $f(x) = \frac{6}{\sqrt{x+13}}$ is not continuous at $x=-13$ . In fact, the function is not defined at $x=-13$ , so it is not possible to determine the left-hand or right-hand limit of the function at this point.

The final answer is that the function $f(x) = \frac{6}{\sqrt{x+13}}$ is continuous for all values of $x$ except $x=-13$ . Therefore, the correct answer is:

A. $x \ \textless \ 13$ is incorrect.
B. $x \ \textless \ 0$ is incorrect.
C. $x \ \textgreater \ -13$ is correct.
D. $x \ \textgreater \ 0$ is incorrect.

The function $f(x) = \frac{6}{\sqrt{x+13}}$ is a rational function with a square root in the denominator. The square root of a negative number is not defined in the real number system, so we need to be careful when dealing with negative values of $x+13$ . Additionally, the function is not defined when the denominator is equal to zero, i.e., when $x+13=0$ .

In conclusion, the function $f(x) = \frac{6}{\sqrt{x+13}}$ is continuous for all values of $x$ except $x=-13$ . Therefore, the correct answer is C. $x \ \textgreater \ -13$ .
Determine the Values of $x$ for Which the Function $f(x) = \frac{6}{\sqrt{x+13}}$ is Continuous: Q&A

In our previous article, we determined that the function $f(x) = \frac{6}{\sqrt{x+13}}$ is continuous for all values of $x$ except $x=-13$ . In this article, we will answer some frequently asked questions about the function and its continuity.

Q: What is the domain of the function $f(x) = \frac{6}{\sqrt{x+13}}$ ?

A: The domain of the function $f(x) = \frac{6}{\sqrt{x+13}}$ is all real numbers $x$ such that $x \geq -13$ . This is because the square root of a negative number is not defined in the real number system, and the function is not defined when the denominator is equal to zero.

Q: Why is the function $f(x) = \frac{6}{\sqrt{x+13}}$ not continuous at $x=-13$ ?

A: The function $f(x) = \frac{6}{\sqrt{x+13}}$ is not continuous at $x=-13$ because the left-hand and right-hand limits of the function do not exist at this point. This is because the square root of a negative number is not defined in the real number system.

Q: Can we simplify the function $f(x) = \frac{6}{\sqrt{x+13}}$ to make it easier to work with?

A: Yes, we can simplify the function $f(x) = \frac{6}{\sqrt{x+13}}$ by rationalizing the denominator. This involves multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x+13}$ . This will eliminate the square root in the denominator and make the function easier to work with.

Q: How do we rationalize the denominator of the function $f(x) = \frac{6}{\sqrt{x+13}}$ ?

A: To rationalize the denominator of the function $f(x) = \frac{6}{\sqrt{x+13}}$ , we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x+13}$ . This gives us:

f(x) = \frac{6}{\sqrt{x+13}} \cdot \frac{\sqrt{x+13}}{\sqrt{x+13}} = \frac{6\sqrt{x+13}}{x+13}

Q: What is the simplified form of the function $f(x) = \frac{6}{\sqrt{x+13}}$ ?

A: The simplified form of the function $f(x) = \frac{6}{\sqrt{x+13}}$ is $f(x) = \frac{6\sqrt{x+13}}{x+13}$ .

Q: Is the function $f(x) = \frac{6\sqrt{x+13}}{x+13}$ continuous for all values of $x$ ?

A: Yes, the function $f(x) = \frac{6\sqrt{x+13}}{x+13}$ is continuous for all values of $x$ except $x=-13$ . This is because the function is defined for all values of $x$ except $x=-13$ , and the limit of the function as $x$ approaches $-13$ exists.

In conclusion, the function $f(x) = \frac{6}{\sqrt{x+13}}$ is continuous for all values of $x$ except $x=-13$ . We can simplify the function by rationalizing the denominator, which gives us the simplified form $f(x) = \frac{6\sqrt{x+13}}{x+13}$ . This function is continuous for all values of $x$ except $x=-13$ .

The final answer is that the function $f(x) = \frac{6}{\sqrt{x+13}}$ is continuous for all values of $x$ except $x=-13$ . The simplified form of the function is $f(x) = \frac{6\sqrt{x+13}}{x+13}$ , and this function is continuous for all values of $x$ except $x=-13$ .