I had one of those weeks this week where it felt like I got a lot of work done yet didn’t make much progress in terms of completing M.P. I ended up writing ~40 pages of notes but only made it 4% of way further through the unit Analyzing Functions. 😞 That being said, I started to get much better grasp on Extreme Mean Theorem and began to intuitively understand how it works which I’m happy about! I also got a better feel for derivatives, in general, and got lot of practice working through relatively tough algebra. Overall, although I didn’t make much progress getting through the unit, I’m happy with my effort I put in this week and felt like I learned a lot.

I spent all of this week working on and understanding the differences between “Relative” extrema and “Local” extrema. My understanding of the difference between each type of extrema is that relative extrema look at all the critical points on a function (a.k.a. the points on a the function where the slope equals 0 or the function is undefined) and label/consider them as extrema, whereas local extrema look at a specific interval along a function and check for critical points within that interval AND ALSO take into consideration the end points of the interval to see if the end points are what’re known as absolute minimum points or absolute maximum points (i.e. if the y-value at those points if the greatest- or least-value when comparing them to the y-value of the critical points within that interval). Simple, right?

Here’s a question from my notes that asked me to find the relative maxima point along the function g(x) = -(x + 1)(x – 1)²:

• Steps to solve:
  • Step 1 – Find the derivative:
    • On the first page you’ll see that I found the derivative of g(x) which equals g’(x) = -3x² + 2x + 1
    • Something worth mentioning is that I had to use the chain rule here, then expand the terms (not sure if I’m saying that right) and then add them all together to come up with the derivative.
  • Step 2 – Find the 0’s:
    • On the right side of the second page you can see some funny looking “X”s with numbers above, below, and beside them. This was me trying to factor the derivative -3x² + 2x + 1. To be honest, I forgot how to factor a trinomial with a leading coefficient (in this case, -3) and had to look it up. After looking it up, I remembered I had to find two numbers that multiplied together to equal the product of the coefficients on the first and last terms in the trinomial (top of the X) and add together to equal the second coefficient of the trinomial (bottom of the X). These numbers ended up being 3 and -1.
    • Using the numbers 3x and -1x in place of 2x, I was then able to factor the trinomial to get (-3x – 1) and (x – 1). At the bottom of the second page you can see the algebra I did to conclude that the 0’s were at x = -1/3 and x = 1.
  • Step 3 – Find out where derivative is (+) and (-):
    • On the third page you can see that I made a number line where the critical values x = -1/3 and x = 1 were labelled with a square drawn around them. Since g’(x) = 0 at these x-values, these are the critical points of the function. This means I needed to figure out the slope of the function before and after these points to determine if they were relative maxima or relative minima points.
    • The x-values I used to test if the derivative was (+) or (-) before and after these points were x = -1, x = 0, and x = 2:
      • g’(-1) = -4
        • (Therefore, the function g(x) was decreasing at this point.)
      • g’(0) = 1
        • (Therefore, the function g(x) was increasing at this point.)
      • g’(2) = -7
        • (Therefore, the function g(x) was decreasing at this point.)
  • Step 4 – Conclusion
    • Since the function g(x) is increasing before the critical point x = 1 and decreasing after that point, you can state that x = 1 is a relative maxima point.

As I mentioned above, finding local extrema (i.e. extrema within a specific interval) follows the exact same process above except that you then also have to check the end points of the interval themselves to see if they’re an absolute maxima or minima point:

Lastly, on an unrelated-to-EVT note, I couldn’t remember how/why the derivative of ln(x) = 1/x so I spent a good chunk of Thursday morning re-watching this video and working through the proof:

I still don’t really understand how/why deriving both sides of a y = x equation works. I know how to work through the chain rule, product rule, quotient rule, etc. but it still seems like magic to me, like I don’t really know what’s going on or why it works. I keep reminding myself that this is exactly how I felt as I worked through trigonometry and algebra, however, and I eventually started to figure both subjects out. I’m hoping the same will happen with derivatives!

It dawned on me this morning that I’m now into my third year of working on KA. Thinking of myself as a third-year student is a bit of a confidence-booster, like I’m not a freshman anymore that has no clue what’s going on. This coming week I’m hoping I can get through a handful of exercises in Analyzing Functions(400/1800 M.P.) and get ~50% of the way through the unit. Looking at the number of videos, exercises, and articles I’d have to get through to get to 50%, I think it will be tough but doable. As well, I finally finished summer camps this past week so I’ll now have my mornings free again which should give me a lot more time to work on KA. So no more excuses!! Time to get shit done! 😤💪🏼