I’m happy to say that I finally (FINALLY) got through the Analyzing Functions unit test. 🥵 I passed it on Friday on my fourth attempt and it took me close to 2 hours to get through it. I used Desmos to double check each of my answers which I think was cheating in a way. There were only 1 or 2 questions that I would have gotten wrong if I hadn’t checked Desmos so, even though I feel a bit guilty about it, I feel ok moving on to the next unit, Integrals. I started the latter unit on Saturday but only got through 3 videos in it and I didn’t take any notes. My plan is to start fresh on Integrals this coming week and ‘hit the ground running’ in it, so to speak. 🏃🏻💨

Below are four questions I worked through on the unit test that I thought were notable.

Question 14)

(Note: I got this question wrong because I didn’t check to see if the second derivative was positive and/or negative on either side of the critical points. I inputted x = 0 as a point of inflection but, since the second derivative is positive on either side of x = 0, that was not correct.)

• Step 1 – Find the first and second derivative of f(x) = -x⁷ + 7x⁶:
  • f(x) = -x⁷ + 7x⁶
  • f’(x) = d/dx[-x⁷ + 7x⁶]
    • = -7x⁶ + 42x⁵
  • f’’(x) = d/dx[-7x⁶ + 42x⁵]
    • = -42x⁵ + 210x⁴
    • = -42x⁴(x – 5)
• Step 2 – State 0’s of second derivative:
  • f’’(x) = -42x⁴(x – 5)
    • Therefore, f’’(x) = 0 @: [x = 0] & [x = 5].
• Step 3 – Using test values [x = -1], [x = 1], and [x = 6], determine whether second derivative is positive or negative on either side of the critical points:
  • f’’(-1) = -42(-1)⁴((-1) – 5)
    • = -42(1)(-6)
    • = 252
    • Therefore, the second derivative is positive to the left of [x = 0], a.k.a.  f(x) is concave up to the left of [x = 0].
  • f’’(1) = -42(1)⁴(1 – 5)
    • = -42(1)(-4)
    • = 168
    • Therefore, the second derivative is positive to the right of [x = 0] (and to the left of [x = 5]), a.k.a. f(x)is concave up between [x = 0] and [x = 5].
  • f’’(6) = -42(6)⁴(6 – 5)
    • = -42(1,296)(1)
    • = -54,432(1)
    • = -54,432
    • Therefore, the second derivative is negative to the right of [x = 5], a.k.a. f(x) is concave down to the right of [x = 5].
• Step 4 – Conclusion:
  • Since the second derivative is negative the left of [x = 5] and positive to the right, f(x) has an inflection point at [x = 5]. However, since the second derivative is positive on both the right AND left of [x = 0], there is not an inflection point at f(0). 

Question 15)

(Note: I got this question correct but not by using the ‘proper’ method. I made and labelled the triangle in the correct way and needed to figure out the value of y but I didn’t think to use the formula y = mx + b. I ended up just using test values for x and y and came to the correct solution.)

• Step 1 – Draw out a triangle with a rectangle in it and label each side of the triangle, and the bottom and side of the rectangle:
  • Bottom of Triangle: length = 8; 
    • x-, y-coordinates [(0, 0), (8, 0)]
  • Side of Triangle: length = 10, 
    • x-, y-coordinates [(8, 0), (8, 10)]
  • Bottom of Rectangle: length = 8 – x, 
    • x-, y-coordinates [((8 – x), 0), (8, 0)]
  • Side of Rectangle: length = y
    • x-, y-coordinates [((8 – x), 0), ((8 – x), y)]
• Step 2 – Using point A, (0, 0), and point B, (8, 10), calculate m:
  • m = y₂ – y₁/ x₂ – x₁
    • = 10 – 0/ 8 – 0
    • = 10/8
    • =5/4
• Step 3 – Create function for the area of the rectangle:
  • Area = Length * Width
    • Length = y
    • Width = x = (8 – x)
  • F(x) = y * (8 – x)
• Step 4 – Using the formula  y = mx + b and knowing that the y-intercept occurs at (0, 0), determine formula for y in terms of x:
  • y = mx + b
    • m = 5/4
    • b = 0
  • y = 5x/4 + 0
  • y = 5x/4
• Step 5 – Input y = 5x/4 into function for area:
  • F(x) = y * (8 – x)
    • = 5x/4 * (8 – x)
    • = (40x/4) – 5x²/4
    • = 10x – 5x²/4
• Step 6 – Find derivative of F(x):
  • F(x) = 10x – 5x²/4
  • F’(x) = d/dx[10x – 5x²/4]
    • = 10 – 10x/4
    • = 10 – 5x/2
• Step 7 – Find critical point(s) of F’(x):
  • F’(x) = 0 @:
    • 0  = 10 – 5x/2
    • -10 = -5x/2
    • -20 = -5x
    • 4 = x
• Step 8 – Compare value of F(4) to values of F(0) and F(8) (the end points) to make sure F(4) has the greatest value and therefore results in the greatest area of the rectangle:
  • F(0) = 10(0) – (5/4)(0)²
    • = 0
  • F(8) = 10(8) – (5/4)(8)²
    • = 80 – (5/4)(64)
    • = 80 – 5(16)
    • = 80 – 80
    • = 0

Question 6)

(Note: This is a question I would have gotten wrong if I hadn’t used Desmos to confirm my answer. I found the critical point of the second derivative, [x = 4], and assumed that was the answer without checking whether it was positive/negative on either side of [x = 4]. When I graphed the function and second derivative, I realized the function is continuously concave up meaning that [x = 4] is not a point of inflection.)

• Step 1 – Find the derivative and second derivative of f(x) = (1/4)x⁴ – 4x³ + 24x²:
  • f(x) = (1/4)x⁴ – 4x³ + 24x²
  • f’(x) = d/dx[(1/4)x⁴ – 4x³ + 24x²]
    • = x³ – 12x² + 48x
  • f’’(x) = d/dx[x³ – 12x² + 48x]
    • = 3x² – 24x + 48
    • = 3(x² – 8x + 16)
    • = 3(x – 4)²
• Step 2 – Determine 0’s of f’’(x):
  • f’’(x) = 3(x – 4)²
    • Therefore, f’’(x) = 0 @: [x = 4]
• Step 3 – Check to see if f’’(x) is positive or negative on either side of [x = 4]:
  • f’’(3) = 3(3 – 4)²
    • = 3(-1)²
    • = 3(1)
    • = 3
    • Therefore, f(x) is concave up to the left of [x = 4]
  • f’’(5) = 3(5 – 4)²
    • = 3(1)²
    • = 3(1)
    • = 3
    • Therefore, f(x) is concave up to the left of [x = 4]
• Step 4 – Conclusion: 
  • Since f(x) is positive on either side of x = 4 and there are no other critical points, f(x) is infinitely concave up with no inflection points.

Question 11)

(Note: The reason why I got this question wrong was because I misread it. I thought the function was f(x) = √(4 – x) and didn’t see that there was an x in front of the radical making the function f(x) = x * √(4 – x).)

• Step 1 – Calculate derivative of f(x) = (x)√(4 – x):
  • f(x) = (x)√(4 – x) = a(x) * b(c(x))
    • a(x) = x
      • a’(x) = 1
    • b(x) = x^1/2
      • b’(x) = 1/2√x
    • c(x) = 4 – x 
      • c’(x) = -1
  • f’(x) = d/dx[(x)√(4 – x)] = a’(x)*b(c(x)) + a(x)*b’(c(x))*c’(x)
    • = (1)(√(4 – x)) + (x)(1/2√(4 – x)(-1)
    • = √(4 – x) – x/2√(4 – x)
      • (Note: I find the algebra at this point very hard and had to look it up using the Symbolab calculator. The next step is to multiply the first expression, √(4 – x), by 2√(4 – x)/ 2√(4 – x) so that the denominators in both expressions are equivalent.)
    • = (2√(4 – x)/2√(4 – x)) * √(4 – x) – x/2√(4 – x)
    • = (2 * √(4 – x) * √(4 – x)) / 2√(4 – x) – x/2√(4 – x)
    • = 2(4 – x)/2√(4 – x) – x/2√(4 – x)
    • = (2(4 – x) – x)/2√(4 – x)
    • = 8 – 2x – x/2√(4 – x)
    • = -3x + 8/2√(4 – x)
• Step 2 – Find the 0’s of f’(x):
  • f’(x) = 0 @:
    • 0 = -3x + 8/2√(4 – x)
    • 0 = -3x + 8
    • -8 = -3x
    • x = -8/-3
    • x = ~2.667
• Step 3 – Conclusion:
  • Since f’(x) = 0 @: [x = ~2.667], the only critical point on the function f(x) is @ [x = ~2.667].

If it wasn’t clear from my intro, I’m VERY happy to finally be through Analyzing Functions and moving onto Integrals (0/3,200 M.P.). From the few videos I’ve watched in the unit so far, I was correct (as far as I can tell) in thinking that integrals calculate the area underneath a function and above the x-axis. Sal also mentioned that an integral is a “anti-derivative” which I think means they’re the reverse operation of derivation, similarly to how division is the reverse operation of multiplication. In any case, I’m eager to start learning about integrals although I feel slightly overwhelmed with how big the unit is. Considering how long it took me to get through Analyzing Functions, my goal is to get through Integrals in the next 2.5 months, before the end of 2021. At the rate I’ve been going, it’s definitely looking like it will take me a longer than a year to learn calculus. 🤷🏻‍♂️