Every once in a while I have a week where everything goes smoothly and I don’t struggle to understand any thing that I work through. This was one of those weeks!! 🥳 I managed to get through 10 videos, 3 articles, and 7 exercises this week which means I’m now 53% of the way through the unit Analyzing Functions. I learned what the concavity of a function means which, as I explain below, refers to whether a function’s slope is increasing in the (+) x-direction or decreasing in the (+) x-direction. Learning about concavity gave me a better appreciation/understanding of second derivatives and why they can be useful to calculate. Overall, I would say this was one of my best weeks working through KA. I learned a ton AND made a lot of progress getting M.P. The icing on the cake is that I’m now a certified pre-algebra tutor!! (See bottom for details. 😉)

Before learning about what a functions’ concavity is, on Tuesday I finished up 2 exercises on Extreme Value Theorem. The exercises dealt with finding a local min/max point within a specific interval on a function and also finding a min/max point across the entire domain of a function. Here are examples of each type of question:

• Q. What is the minimum value of h(x) = x³ – 6x² + 8 over the closed interval -1 ≤ x ≤ 6?
  • Step 1 – Find the critical points by taking the derivative of h(x):
    • h(x) = x³ – 6x² + 8
      • h’(x) = d/dx[x³ – 6x² + 8]
        • = 3x² – 12x 
  • Step 2 – Find “0’s” of h’(x):
    • h'(x) = 3x² – 12x
      • = 3x(x – 4)
      • Therefore, 0’s occur at x = 0 and x = 4 since inputting those values into 3x(x – 4) results in the expression equaling 0.
  • Step 3 – Input the “0’s” x-values AND endpoint x-values into original function to see which h(x) is the lowest and therefore the minimum point:
    • h(-1) = (-1)³ – 6(-1)² + 8
      • = –1 – 6(1) + 8
      • = –1 – 6 + 8
      • = 1
    • h(0) = 0³ – 6(0)² + 8
      • = 0 – 6(0) + 8
      • = 0 – 0 + 8
      • = 8
    • h(4) = 4³ – 6(4)² + 8
      • = 64 – 6(16) + 8
      • = 64 – 96 + 8
      • = –24
    • h(6) = 6³ – 6(6)² + 8
      • = 216 – 6(36) + 8
      • = 216 – 216 + 8
      • = 8
  • Therefore, the local min of h(x) = x³ – 6x² + 8 over the closed interval -1 ≤ x ≤ 6 is at x = 4 which equals -24.

• Q. What is the absolute maximum value of g(x) = 3x⁴ + 8x³ + 4 over the entire domain of g?
  • Step 1 – Find the critical points of the function by taking its derivative:
    • g'(x) = d/dx[3x⁴ + 8x³ + 4]
      • = 12x³ + 24x²
  • Step 2 – Find the 0’s of the derivative:
    • g'(x) = 12x³ + 24x²
      • = 12x²(x + 2)
      • Therefore, 0’s occur at x = 0 and x = -2 since inputting those values into 12x²(x + 2) results in the expression equaling 0.
  • Step 3 – Using the derivative and test x-values, find out whether the derivative is positive or negative on either side of the two critical points and in between:
    • g'(–3) = 12(–3)²((–3) + 2)
      • = 12(9)( –1)
      • = –108
        • Therefore, the slope of the function is decreasing at this point and to the left of x = –3.
    • g'(–1) = 12(–1)²((–1) + 2)
      • = 12(1)(1)
      • = 12 
        • Therefore, the slope of the function is increasing at this point and throughout –2 < x < 0.
    • g'(1) = 12(1)²((1) + 2)
      • = 12(1)(3)
      • = 36
        • Therefore, the slope of the function is increasing at this point and to the right of x = 1.
  • Therefore, as you can see from the graph, since the value of the function is increasing in both the (-) and (+) directions along the x-axis, there is no absolute maximum value of this function across its entire domain since it increases to infinity in both directions.

I finished those two exercises on Tuesday and began learning about the concavity of a function which lasted the remainder of the week. As you might imagine from the name “concavity”, this concept refers to the shape of a function where, if it’s an ‘upright’ “U”, it’s referred to as “concave-upwards” and if it’s an ‘upside down’ “∩”, it’s referred to as “concave-downwards”. When the concavity of a function is upwards across a certain interval. it means the function’s slope is increasing when moving from left to right along that interval. The opposite is true when a function’s concavity is downwards. The point at which a function switches from concave-upwards to concave-downwards, or vice versa, is called an “inflection point”:

The way you calculate a function’s concavity is by taking its second derivative. This part is still a bit confusing to me as to why why it works, but if a functions’ second derivative is positive then the concavity is upwards (i.e. the functions’ slope is increasing) and if a functions’ second derivative is negative along a certain interval then the concavity is downwards (i.e. the functions’ slope is decreasing). 

I spent the rest of the week watching videos, reading articles, and going through exercises that all had to do calculating a function’s concavity. Doing this helped me solidify the idea that the derivative of a function indicates the value of the slope at any given point along the function, and the second derivative indicates whether the slope is increasing or decreasing at any point along the x-axis of the function. This concept is something I’ve been struggling to fully get straight in my mind, but it’s been becoming more clear to me over the past few weeks (thank the lawd).

I mentioned last week (and maybe in other weeks, as well?) that derivatives seem like magic to me and that I’ve been having a hard time understanding how/why the sum, product, and chain rule all work. This week I found 3 videos that really helped me to understand how/why these rules work which I’ve copied just below. I still haven’t completely figured it all out but I found these videos which started to make it all more clear and which I thought were very useful:

On a final side note, a few weeks ago I saw an interview that Sal gave about a website he and his team created called www.schoolhouse.world which is a platform that utilizes KA for people to get tutoring and become tutors. I signed up to become a tutor and went through the pre-algebra course challenge, passed it, and am now a certified pre-algebra tutor! My plan is to certify myself in Algebra 1, Algebra 2, Geometry, Trigonometry, Pre-Calculus, and Calculus. I’d like to get certified in all those subjects before the end of the year which, except for maybe Calculus, I think is doable. At that point, I think it’ll help me feel much more confident about the idea of tutoring math to kids which is very motivating and an exciting idea to me!

I only have 5 exercises left to go in Analyzing Functions (960/1800 M.P.), but there are 19 videos and 2 articles remaining so, unless I continue flying through the material as fast as I did this week, it seems likely that it’ll take me the rest of the month to get through the unit. I think it’s also becoming less and less likely that I’ll get through Calc 1 before the end of the year, but I’m still planning on making it happen! I feel like passing Calc 1 would be a great (read: nerdy) way to ring in the New Year. 🥳 🤓