Going by the, “there are 52 weeks in a year” saying, I’m offcically 2 years into using KA! (Although technically that’s not accurate since there are actually 52 weeks and 1 day per year, plus last year was a leap year, so I won’t hit the official 2-year mark until this coming Wednesday.) Reflecting on how much I’ve learned since I began, I’m very proud of how far I’ve come and the direction I’m going. Woo!! 🥳 🎉🎊

This week I didn’t get as far through the unit Analyzing Functions as I was hoping I would, but I did manage to get through a good chunk of material, nonetheless. It took me 2 days to get through 1 exercise dealing with extreme value theorem (EVT) which was disappointing but, at the same time, it gave me a lot of practice doing relatively difficult algebra which I improved at. After that exercise, I spent the rest of the week watching 5 videos and doing 2 exercises on what are known as Relative Extrema which have to do with determining if there’s a critical max/min point on a function by using the function’s derivative. It seems like derivatives are slowly starting to make more sense to me which I’m very happy about but, that being said, I think I still have a long way to go before I fully understand them completely.

Understanding how EVT works isn’t too tough for me now but the algebra I need to use in some of the questions to solve for the answer can be pretty difficult. The questions I worked through in the first exercise this week had me either use EVT to find the critical points of a function or figure out where the slope of a function was increasing or decreasing between a certain interval. Here’s an example of each type of question:

• Q. 1) Where does the function f(x) = e^{x^2 + 2x} have critical points?
  • (Note: What this question is asking me to do is figure out where the function is undefined [i.e. where the numerator is being divided by 0] and/or where the “0’s” are [i.e. where the numerator equals 0]. These would be the critical points of the function.)

• Step 1 – Find the derivative of f(x) = e^{x^2 + 2x}:
  • f(x) = e^{x^2 + 2x} = a(b(x) + c(x))
    • a(x) = e^x
    • a’(x) = e^x
    • b(x) = x²
    • b’(x) = 2x
    • c(x) = 2x
    • c’(x) = 2
  • f’(x) = a’(b(x) + c(x)) * (b’(x) + c’(x))
    • = e^{x^2 + 2x} * (2x + 2)
      • (Note: At this step you need to factor a 2 out of the term (2x + 2) which is what tripped me up on this question.)
    • = 2e^{x^2 + 2x} * (x + 1)
• Step 2 – Figure out if/where the derivative is either undefined or equals 0:
  • The function is never undefined since it’s not being divided by 0.
  • The term 2e^{x^2 + 2x} will never equal 0 since the variable x is an exponent on e and there is no number that can be used as a power to a non-zero number that will make it’s value equal to 0.
    • (Note: If you raise e to the power of 0, e⁰, it equals 1.)
  • The term (x + 1) equals 0 when x = -1:
    • x + 1 = 0
    • x = -1
  • Therefore, the only critical point on the function f(x) = e^{x^2 + 2x} is at x = -1.

The reason why I find EVT easy to understand it because if you input the function into Desmos you can see it’s graph and you simply look for where the function has a min/max point or an asymptote and those are the critical values: 

• Q. 2) Given the derivative g’(x) = x²/(x – 2)³, select the intervals where  the function g(x) is increasing.
  • (Note: The key here is to figure out where the derivative g’(x) is undefined and where it’s 0’s are. Once you’ve figure those out, you split the function up into intervals at the undefined points and where g(x) = 0 and then figure out which intervals are positive, i.e. where the slope of g(x) is positive.)

• Step 1 – Determine where g’(x) is undefined and where it equals 0:
  • g’(2) = (2)²/((2) – 2)³
    • = 4/(0)³
    • = Undefined
  • g’(0) = 0²/(0 – 2)³
    • = 0
  • Therefore, the function is undefined at x = 2 and equals 0 at x = 0.
• Step 2 – Determine what the intervals are and decide on an x-value to use that’s inside each interval to check to see if the interval is positive or negative:
  • Critical points = 0 & 2.
  • Intervals = [-∞, 0], [0, 2], and [2, ∞]
  • X-values to check whether each interval is positive or negative:
    • x = -1
    • x = 1
    • x = 3
• Step 3 – Input the x-values into the derivative to figure out if the derivative is positive or negative:
  • f’(-1) = (-1)²/((-1) – 2)³
    • = 1/(-3)³
    • = -1/27
    • Therefore, since f(-1) is negative we can say that g(x) is NOT increasing along the interval [-∞, 0].
  • f’(1) = (1)²/((1) – 2)³
    • = 1/(-1)³
    • = -1
    • Therefore, since f(1) is negative we can say that g(x) is NOT increasing along the interval [0, 2].
  • f’(3) = (3)²/((3) – 2)³
    • = 9/(1)³
    • = 9
    • Therefore, since f(3) is positive we can say that g(x) IS increasing along the interval [2, ∞].

On Thursday I began working on the next section which dealt with what Sal referred to as “extrema” which I’m pretty sure is just a fancy word for a min/max point of a function. This section was titled Relative (Local) Extrema and went through the math behind determining whether a point on a function that is either undefined or equals 0 is considered a critical point and, if so, if it’s a maximum or minimum point. My layman’s explanation of what was taught is, if the derivative of a function changes from either (+) to (-) or (-) to (+) then it is a min/max point, a.k.a. it’s an extreme point:

If you can decipher my terrible hand writing, you can see that there are 2 points, x₀ and x₁, that are maximum and minimum points, respectively, since their slopes change from (+) to (-) and to (-) to (+), respectively. I got through 5 videos and 2 exercises in this section where all the questions asked me to determine the intervals on functions between the undefined points and where the function equaled 0 and, from there, figure out if the derivatives switched signs at those points to determine if they were max/min points. I’m not going to go through the math, but here are 2 pages from my notes where I practiced solving these types of questions:

On a final side note, I was given a question that required me to factor out a negative exponent from an expression and I was completely stumped on how to do it. I eventually figured it out but had to look it up on the Symbolab calculator in order to figure it out. Here’s the start of the question and my notes on how to factor it:

(Note: In order to find the relative maximum point on this function, I needed to find the derivative of h(x), then find where the function had undefined, maximum and/or minimum points and figure out if the derivative went from positive to negative along those intervals.)

• h(x) = x² + 1/x² = a(x) + b(x)
• h’(x) = a’(x) + b’(x) = 2x + (-2x^-3)
  • (This is the part that I was stuck on. I knew I needed to factor out the 2x from both terms but didn’t realize I needed to add a x³x^-3 to the first term of the equation. You’re able to do this because x³x^-3 = (x * x * x)/(x * x * x) = 1.)
  • = 2x⁴x^-3 + (-2x^-3)
  • = 2x⁴x^-3 – 2x^-3
  • = 2x^-3(x⁴ – 1)
  • = 2(x⁴ – 1)/x³
  • The derivative h’(x) is undefined when x = 0 and equals 0 when x = 1.

(Note: I’m not going to bother solving the full question here, but to do so I then needed to figure out the (+) and (-) signs of the intervals [-∞,0], [0, 1], and [1, ∞] and figure out if they switched from (+) to (-) going from left to right along the x-axis.)

It’s hard to believe I’m going to be starting Week 105 this coming week. I’m now 18% of the way through the unit Analyzing Functions (320/1800 M.P.) so I think I’m on track to get through the unit before the end of September. When I hit the official 2-year mark this coming Wednesday, I’ll essentially have gone from the beginning of Grade 6 math to the middle of first year university math. (To be fair, I started using KA in the summer before Sept. 2^nd, 2019 so, if I factor that in, I actually started from Grade 3 math!) My goal is to get through all of the math sections on KA around the 3.5-year mark. I’m also hoping to start a math tutoring program at my squash club for the students I work with and will hopefully figure out a way to use KA as the platform to tutor them on. I figure if it’s worked this well for me, I don’t see any reason why it wouldn’t work for them!