I had another relatively disappointing week this week. 😔 I got one question wrong on my first attempt at the Analyzing Functions unit test and scored 17/18. I’m now halfway through my second attempt at the test but have already gotten 2 questions wrong so I’ll need to do it at least one more time next week. The silver lining is that all of the questions I got wrong were questions I could (and probably should) have gotten correct. As well, I was quite happy with how well I understood all of the questions and my comprehension of what I was being asked (for the most part).

Since all I did this week was work through the unit test, here are the three questions I got wrong, and 1 other question that I actually got correct (Question 16) but thought I’d mention simply because I was happy with how well I understood it:

Question 4:

(Note: I got this question wrong because I didn’t think the derivative that was calculated in Step 1 was correct. When I tried solving the derivative myself, I got to the equation 2x + 2y² + 4xy * dy/dx = 0 but then forgot that I needed to isolate dy/dx. In retrospect, this was a stupid mistake to make considering that the answer given in Step 1 literally says dy/dx = … which implies you need to isolate dy/dx. 🤦🏻‍♂️)

• Step 1 – Using implicit differentiation, find the derivative of x² + 2xy² = 25:
  • d/dx[x² + 2xy²] = d/dx[25]
    • 2x + 2((1)(y²) + (x)(2y)(dy/dx)) = 0
    • 2x + 2y² + 4xy * dy/dx = 0
    • 4xy * dy/dx = –2x – 2y²
    • dy/dx = (–2x – 2y²)/4xy
    • dy/dx = (–x – y²)/2xy
• Step 2 – Note that, based off 2^nd equation, either x = 0 or y = 0:
  • 2xy = 0
    • Therefore, for equation to equal 0 either x = 0 or y = 0 or both.
• Step 3 – Substitute both x = 0 and y = 0 into first equation:
  • x² + 2xy² = 25
    • @ x = 0:
      • 0² + 2(0)y² = 25
      • 0 + 0 ≠ 25
    • @ y = 0:
      • x² + 2x(0)² = 25
      • x² + 0 = 25
      • √x² = √25
      • x = (+/-) 5
• Step 4 – Conclusion:
  • “Step 3” in the question states that, “x = -5, x = 0, and x = 5” however x ONLY equals -5 and 5 and not 0 so “Step 3” is incorrect.

(Note: here’s a photo from my notes of me properly solving the derivative for the given function which is what I screwed up initially.)

Question 16:

• Step 1 – Solve 1^st derivative of g(x):
  • g(x) = x⁴ + 4x³ – 48x² + 6
  • g’(x) = d/dx[x⁴ + 4x³ – 48x² + 6]
    • = 4x³ + 12x² – 96x
• Step 2 – Solve 2^nd derivative of g(x):
  • g”(x) = d/dx[4x³ + 12x² – 96x]
    • = 12x² + 24x – 96
    • = 12(x² + 2x – 8)
    • = 12(x + 4)(x – 2)
• Step 3 – find critical point of 2^nd derivative:
  • g”(x) = 12(x + 4)(x – 2)
    • Therefore, g’’(x) = 0 @:
      • x = –4 and x = 2
• Step 4 – Since critical points of 2^nd derivative are @ x = –4 and x = 2, use testing points @ x = –5, x = 0, and x = 3 to determine where 2^nd derivative is positive and/or negative and, therefore, concave up and/or concave down:
  •  g”(-5) = 12(-5 + 4)(-5 – 2)
    • = 12(-1)(-7)
    • = 12(7)
    • = 84
    • Therefore, g’’(-5) = 84, a positive number, meaning g(-5) is concave up.
  • g’’(0) = 12(0 + 4)(0 – 2)
    • = 12(4)(-2)
    • = 12(-8)
    • = -96
    • Therefore, g’’(0) = -96, a negative number, meaning g(0) is concave down.
  • g’’(3) = 12(3 + 4)(3 – 2)
    • = 12(7)(1)
    • = 12(7)
    • = 84
    • Therefore, g’’(3) = 84, a positive number, meaning g(3) is concave up.
• Step 4 – Conclusion:
  • The only test point where the 2^nd derivative was negative (i.e. concave down) occurred @ x = 0 and therefore, given the 4 answers to choose from, the interval in which g(x) is concave down is between (–4, 2) only.

Question 6 (2^nd U.T. attempt):

When I worked through this question, I first thought I had to find the 0’s of g’(x) = 3sin(x) + ln(x) which I found pretty difficult to figure out and don’t think I actually managed to solve. I realized later that it said to simply use a graphing calculator so I plugged the function into Desmos and was shown the graph in the second screen shot but for some reason I thought I was looking at g(x) and, since there are 2 critical points on the function, I thought the answer was Two. Of course, since the graph is of g’(x), the graph indicates that on the function g(x) there’s a single critical point where g’(x) = 0 which occurs at x = ~3.1. 😡

Question 12 (2^nd U.T. attempt):

I wasn’t 100% sure what I had to do for this question but assumed I needed to find out what the slope was between the points g(100) and g(101). I did the calculation which showed that m was indeed ½ which made me thing that g’(x) between x = 100 and x = 101 would have equalled ½ somewhere within that interval. My mistake in this question was thinking that a continuous function means it’s also differentiable, which is not true. After reviewing my notes, I remembered that a differentiable function IS continuous but a continuous function ISN’T necessarily differentiable. As an example, the function in the screen shot below is continuous (i.e. there are no line discontinuities) however, since there’s a “sharp turn”, it’s not differentiable (i.e. at the sharp turn, you cannot find it’s derivative, a.k.a. there is no specific tangent line at that point):

My hope this coming week is that I get through this unit test at least by Friday so that I can wrap up Analyzing Functions (1780/1800 M.P.) and move onto Integrals (0/3200 M.P.). I’m really looking forward to learning about integrals because, as I said in the end of my last post, from what I understand they’re a core part of calculus and I feel like once I get a handle on them I’ll feel like I’m over the hump of understanding calculus. In other words, I’ll finally be a full-fledged nerd. 🤓