Weekly Math Problem #77

Asymptotes. Oh man...I was "this" 🤏🏿 close to skipping this week's problem. Nonetheless, I pressed on. I was scrolling through a College Algebra Study Guide (from my personal archives) when I came across the question you will see below. When I read it, it thought to myself, "Do I remember how to find asymptotes?". 🤔 Then I recalled that there are some rules to remember when it comes to finding any existing asymptotes for rational functions. So, this topic is the focus of this week's WMP.

Check out WMP #11 where I did some previous work with asymptotes.

To solve this week's problem in completion, you need to recall the following math skills and information:

        ✔️     Rules for finding asymptotes 
        ✔️     How to solve linear equations

            

WMP #77 says ...

Happy solving!

Check back on Saturday, April 2nd for the solution, which will be posted below ⬇️.

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✏️📓 Solution Time! 📓✏️

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Let's start with the HA (horizontal asymptote). There are two ways you can find the HA. One way is to pay attention to the degrees of the numerator and denominator, then follow the rule. In our case, the degree of the numerator and denominator are the same (each is degree one ---> linear). When the degrees are the same, then the value for the equation of the line is the ratio of the leading coefficients.

Alternatively, you can evaluate the function at a very large value...say 100 or more. (The larger, the better, as long as your machine can handle it.) The result will yield the value or an an approximate value for the horizontal asymptote. Remember, the equation for a horizontal line is y = , f(x) = , etc.

As for the VA (vertical asymptote), all you need to do is set the denominator equal to zero and solve for x. Your solution is the equation of the VA.

No oblique asymptotes exist because our function doesn't fit the criteria for having any.

Here is a graph of the function:

 

** This plot was generated using Geogebra.org's calculator. **

▪️ Were you able to find the asymptotes?

▪️ Let me know what you thought about this week's problem in the comments section. 

Thank you for solving with me this week. ✏️
We're on to WMP! #78. 🤓

Cheers!

The Younge Lady

Weekly Math Problem #72

Limits. I found this week's problem in my personal archives. 😁 I'm inspired by students I'll be working with that are learning limits. I can recall learning the concept of what a limit is and how to evaluate the limit of various types of functions. Honestly, I learned limits more after my Calculus 1 course as tutor than when I was enrolled in the course. Repetition really is the key 🗝 for me. Seriously, if I don't use it, I can definitely lose it. **Whispering** "This is why I started my blog." 😊

To solve this week's problem in completion, you need to recall the following math skills:

        ✔️     Substitution        
        ✔️     Simplifying rational expressions
        ✔️     Web Resource: Limits (Evaluating)

             

WMP! #72 want us to...

Happy solving!

Check back on Saturday, February 5th for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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To begin the process of finding the limit, we'll use the substitution method. 

Unfortunately, the substitution method yields an indeterminate form expression. What does that mean? 🤷🏿‍♀️ This means that the method used doesn't tell us whether or not the limit exists. Another method for evaluating the limit needs to be used to determine whether or not the limit exists.

The numerator of the expression is a cubic polynomial that can be factored. The form is a difference of cubes. When the cubic expression is factored, you will see that one of the factors matches the expression in the denominator. **Yes!** So, we'll simplify the expression. This time, the substitution method will work.

Nice. The limit does exist, and it's equal to 3. 

Another method that can be used to evaluate a limit that yields and indeterminate form is L'Hôpital's Rule. It involves using derivatives and yields the same results. Take a look...

I didn't do it here, but you can graph the original expression and the simplified expression to see what their graphs look like. Then verify that as x approaches 1 from the left and right, the output value is 3.

▪️ Were you able to find the limit?
▪️ Did you do something else to find the limit? If so, please share. (No judgment.)

▪️ Let me know what you thought about this week's problem in the comments section. 

Thank you for solving with me this week. 😊
Up next...WMP! #73. 

Cheers!

The Younge Lady

Weekly Math Problem #62

Improper Integral. Ever so often, I find my way back to calculus, and this week it's via an improper integral. By my recollection, the technique is okay. The type of function you're dealing with can definitely be the cause that the problem is annoying to do. Let's just hope this week's problem isn't annoying to you. 😬 

To solve this week's problem in completion, you need to recall the following math skills:

        ✔️     How to solve an improper integral
        ✔️     Methods of integration

     

WMP! #62 says to...

Happy solving!

Check back on Saturday, November 6th for the solution, which will be posted below ⬇️.

Shameless 🔌 Plug: Follow me on Instagram @TheYoungeLady
Buy Me a ☕️ Coffee: TheYoungeLady ( I'm gonna need it this year. 😆 )

✏️📓 Solution Time! 📓✏️

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Alright...let's get into this problem! If you looked up integral tables and used a calculator, that's good. I did too! I wasn't expecting anyone to have the information needed to solve this problem memorized. I didn't have it memorized, but I felt confident that I knew how to use the assistance.

I started this problem by using an integral table. Why? Well, I saw that the function was rational and knew that most, if not all, integral tables have some rational functions on them. When I saw what I needed, I rewrote the function in a way that help me begin solving the problem.

With the problem rewritten, I could see that I needed to invoke the substitution method to keep going. Since the substitution method uses a new variable, I also adjusted the lower bound of integration to match the new variable.

Once that was done, I could now move forward and integrate.

Since this is a definite integral, we need to evaluate! Evaluating the second term, which comes from the lower bound was simple. If you're familiar with this special value, then you can simple write the value. You can also use a calculator like I did. 😁 It's the first term, that requires a bit more work. Infinity isn't a value to plug in, so I used my calculator, again, to see if the function converges as my variable approaches infinity. It does!!!

Now that I have the values I need, I can plug them in and simplify.

I don't want to leave my answer like that, so I did a little "clean-up" by rationalizing the denominator.

Here is an image of the function. Keep in mind that the variable is approaching infinity. Even though, the shape doesn't end, the area under the curve converges to value above. 

**This plot was created using Geogebra's Graphing Calculator. Click image to enlarge.

▪️ Were you able to solve the improper integral??

▪️ Let me know what you thought about this week's problem in the comments section. 

Thanks for solving with me this week!

WMP! #63. is up next. 👩🏿‍🏫

Cheers!

The Younge Lady