Determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether

Determine all horizontal and vertical asymptotes. For each...

Key Concepts

Vertical Asymptote Analysis

Vertical asymptotes occur where a function becomes unbounded as the input approaches a specific finite value. In rational expressions and other functions, these are typically found by identifying values of the variable that cause the denominator to vanish (without a corresponding cancellation with the numerator) and then examining the limits from both sides. This type of analysis is crucial when determining the function’s behavior near points of discontinuity.

Horizontal Asymptote Analysis

Horizontal asymptotes describe the behavior of a function as the independent variable approaches positive or negative infinity. They are found by taking the limits of the function at infinity, revealing if the function stabilizes at a constant value. Understanding how to determine horizontal asymptotes is vital for analyzing the end behavior of functions.

Domain of Logarithmic Functions

The logarithmic function is defined only for positive arguments, which immediately imposes restrictions on the domain of any function that includes logarithms. Recognizing and applying these domain restrictions ensures that subsequent limit and asymptotic analyses are performed on the appropriate set of input values.

One-Sided Limit Analysis

One-sided limits involve examining the behavior of a function as the variable approaches a particular value from either the left (negative direction) or the right (positive direction). This concept is especially important when analyzing vertical asymptotes, as the function’s tendency to approach positive or negative infinity may differ based on the direction of approach.

Behavior of Logarithmic Functions near Critical Points

The natural logarithm function exhibits characteristic behavior near critical points; it tends to negative infinity as its argument approaches zero from the positive side and increases slowly for large arguments. Understanding these behaviors is key in analyzing limits and asymptotic behavior when logarithms are involved, particularly when the argument of the logarithm approaches values near zero or infinity.

Transcript

00:01 All right, we've got a question here that describes the function of x to be equal to x plus two, excuse me, the ln of x plus two over the ln of x squared plus three x plus three.

00:20 Okay, first we want to determine what our values of x have to be.

00:24 We know that we can't have a zero, excuse me, an ln of zero.

00:30 So we would have our x plus two greater than zero.

00:34 And we would say, well, x has to be greater than negative 2.

00:39 So we can already imagine a graph, point negative 2 here.

00:47 We can say that our region would be somewhere along the right side of negative 2.

00:53 Now, in order to determine our asymptotes, we're going to take this polynomial, and we're going to determine what our values of x would be.

01:06 And we would set this equal to not equal to 1, because you know first of all we know it has to be let me clarify this so this has to be greater than zero because we know we can't have the ln of zero but then also it has to be greater than 1 because the ln of 1 is 0 and then we would have an undefined function so what we do is we solve for this by taking the factorial excuse mean by factoring this as simply as we can.

01:59 So we would say, well, x plus 2x plus x plus 3.

02:07 And we could factor this out by saying, okay, we could have, we could subtract one from both sides.

02:29 So we would have x plus 2 greater than 0.

02:36 Okay, and now here we would say that x equal to negative 2.

02:42 So one of our vertical asymptopes, and we can see say it's negative 2.

02:47 We would then take the limit as x approaches negative 2 for the function of x, which is x approaches negative 2, ln x plus 2, ln of x squared plus 3x plus 3.

03:13 Now we know that our region here, we said it's approaching negative 2.

03:19 So if we put our x approaches 2 from the positive side, side, and we could solve for whether it's going to the negative infinity or to the positive.

03:30 So we plug in negative 2 in for our x, and we're going to get ln 0 over ln of negative 2 squared is 4 plus negative 6 is negative 2.

03:49 We're going to have ln 1.

03:51 Therefore, we have an infinity or a 0.

03:56 We have a 0 at the denominator.

03:58 So we would have this be equal to ln of x plus 2 over 0, which approaches a positive infinity.

04:20 All right...

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