I didn’t get very far through the unit Analyzing Functions this week. In total I only made it through 4 videos and 4 exercises. I finished the section on Mean Value Theorem (MVT) and started working through the following section which dealt with Extreme Value Theorem (EVT). There were only 3 videos on EVT which I finished watching but wasn’t able to finish the 1 exercise on EVT before the end of the week. Between the 5 exercises I worked on this week in total, most of what I did involved algebra, which I usually enjoy, but it was some of the hardest algebra I’ve ever had to do so I struggled with it which was frustrating. As always, I’m sure I’ll figure it all out with a bit more practice. I started getting the hang of it by the end of the week so I’m confident I’ll figure it all out soon enough. 😤

I spent Tuesday and Wednesday wrapping up the section on MVT. This photo does a pretty good job explaining MVT:

My layman’s explanation of MVT is it states that when looking at a continuous function between a specific interval ([x₁, x₂] or [a, b]) the slope between the 2 points of the interval, m, will be equal to the derivative of at least one point along the function (f'(x₃) or f'(c)).

It took me awhile to understand how MVT works. After going through all the KA videos and attempting the exercises, I watched this video which really helped me to understand how it works:

The key thing to remember is to set up the equations for MVT as: 

• f’(c) = f(b) – f(a)/b – a
  • “The derivative (a.k.a. the slope) at f’(c) is the same value as the slope between the two points f(a) and f(b). 

Here’s an example question I worked through:

(Note: The question states that h(x) = (3x – 3)^1/2 and that there’s a number, c, that satisfies MVT which implies the function is continuous and differentiable. The question is asking you to find where the derivative of h(x) equals the slope between the 2 points x = 4 and x = 13.)

• Step 1 – Set up equation:
  • h’(c) = m = (y₂ – y₁)/(x₂ – x₁) = (h(13) – h(4))/(13 – 4)
• Step 2 – Solve for h(x₁) and h(x₂):
  • h(4) = (3(4) – 3)^1/2
    • = (12 – 3)^1/2
    • = 9^1/2
    • = 3
  • h(13) = (3(13) – 3)^1/2
    • = (39 – 3)^1/2
    • = 36^1/2
    • = 6
• Step 3 – Solve for m:
  • m = (h(13) – h(4))/(x₂ – x₁)
    • = (6 – 3)/(13 – 4) 
    • = 3/9
    • = 1/3
• Step 4 – Find derivative of h(x)
  • h(x) = (3x – 3)^1/2  = (a(b(x))
    • a(x) = x^1/2
    • a’(x) = 1/(2x^1/2)
    • b(x) = 3x – 3
    • b’(x) = 3
  • h’(x) = d/dx[(3x – 3)^1/2] = d/dx[(a(b(x))] = a’(b(x)) * b’(x)
    • = 1/(2(3x – 3)^1/2) * 3
    • = 3/(2(3x – 3)^1/2)
• Step 5 – Solve for h’(c) = m
  • 3/(2(3(c) – 3)^1/2) = 1/3
  • 3(3) = 2(3(c) – 3)^1/2
  • 9 = 2(3(c) – 3)^1/2
  • 9/2 = (3(c) – 3)^1/2
  • (9/2)² = 3(c) – 3
  • 81/4 = 3(c) – 3 
  • 81/4 + 12/4 = 3(c)
  • 93/4 = 3(c)
  • 93/4/3 = c
  • 7.75 = c
• (Note: This means that on the interval of [4, 13] on the function h(x) = (3x – 3)^1/2, the derivative at x = 7.75, a.k.a. the slop at x = 7.75, is the same as the slope between the 2 points x = 4 and x = 13.)

I began working on EVT on Thursday which I spent the rest of the week working on. There were 3 videos that I had to get through which took me until Friday morning and 1 exercise which I wasn’t able to finish before the end of Saturday. I’m starting to understand EVT but I’m still a bit confused by it, plus the algebra I had to do in the exercise was some of the hardest algebra I’ve had to do up to this point which is why I didn’t get through it. Here are a few of my notes that explain EVT:

(Note: Usually I would rewrite these type of notes before posting them since they’re so messy but I’m running late to so I’m not going to bother! I apologize for the terrible hand writing.)

The gist of EVT is that, when looking at a closed interval on a continuous function, there will always be a minimum point and a maximum point. The way you state EVT mathematically (as shown at the top of the second photo) is:

• ∃ c, d ∈ [a, b]: f(c) ≤ f(x) ≤ f(d)
  • “There exists (∃) an x-value (c, d) that’s a member of (∈) the interval [a, b] where (:) f(c), a minimum value, is less than or equal to (≤) the entire function of f(x) which is less than or equal to(≤) f(d), a maximum value.”

This seemed arbitrary to me at first but then I learned about Critical Points which made EVT seem more useful:

In the function at the top of the first photo, you can see a few labelled points on the graph:

• Global Maximum:
  • The greatest y-value on the function for the span of the entire function.
• Global Minimum:
  • The lowest y-value on the function for the span of the entire function.
• Local Maximum:
  • The greatest y-value on the function for a specific interval on the function.
• Local Minimum:
  • The lowest y-value on the function for a specific interval on the function.

Global maximum and minimum, and local maximum and minimum points are all referred to as critical points. Part of what this means is, at these points, the derivative of the function is either equal to 0 or it’s undefined. There are 2 important caveats:

1. Critical points cannot be at the end points of the interval (as shown in the bottom of the second photo – not sure why exactly.)
2. If the OG function is not defined (or possibly equal to 0 – again not sure) at a certain point, the derivative at that specific point will not be considered a cortical point if it is also undefined (or perhaps equal to 0).

As I’ve said a few times, I’m still pretty confused about how this all works. Each question in the exercise had me find the derivative of a given function and then figure out if the derivative equaled 0 or was undefined at any point and, if it was, state that those points were critical point. Here’s an example:

• Step 1 – Find the derivative of e^2x/x – 3:
  • h’(x) = d/dx[e^2x/x – 3] = d/dx[a(b(x))/c(x)]
    • = (a’(b(x)) * b’(x) * c(x) – (a(b(x)) * c’(x))/[c(x)]²
      • (Note: This is a combination of the chain rule and quotient rule.)
    • = (e^2x * 2 * (x – 3) – e^2x * (1))/(x – 3)²
    • = ((2e^2x * (x – 3)) – (e^2x * (1)))/(x – 3)²
    • = e^2x(2(x – 3) – (1 * 1))/(x – 3)²
    • = e^2x(2x – 6 – 1)/(x – 3)²
    • = e^2x(2x – 7)/(x – 3)²
• Step 2 – determine if h’(x) equals 0 or is undefined:
  • e^2x(2x – 7)/(x – 3)² = 0
  • e^2x(2x – 7) = 0 * (x – 3)²
  • e^2x(2x – 7) = 0
    • (Note: I don’t know how to do this algebra, but when x = 3.5 the equation equals 0 because e^2(3.5)(2(3.5) – 7) = e^2(3.5)(7 – 7) = e^2(3.5)(0) = 0.)
  • … x = 3.5
• As shown in the second screens shot, h’(x) is also undefined at x = 3 but since the OG function is also undefined at x = 3 it does not count as a critical point.

Just from writing out the math above, it’s pretty obvious to me that I don’t have a great grasp on what I’m doing or on the algebra I need to solve these types of questions. I will keep at it though until I figure it out!

This coming week I’d like to get through a section and a half of the unit Analyzing Functions (160/1800 M.P.) which is approximately 6 videos and 3 exercises. Even if I manage that, it looks like it’s going to be tough for me to get through the entire unit before the end of September. Hopefully this coming week isn’t as tough as this past week but, somehow, I doubt it. 🥵