Solving for Critical Points Using a Derivative

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I would like to know how the equation was solved. I was not able to solve the equation manually even after looking at the answers:

derivative = "3x^2 - 2x"
critical_points = [0, 2/3]
rel_min = [2/3]
rel_max = [0]

The instructions say to equate 3x**2 - 2*x to 0. I assume it was because I need a point with a slope of zero.
How were those numbers listed in the answer section solved to be the derivative’s critical points? The farthest I got solving 3x**2 - 2*x = 0 was putting either constant on the either side. I get stuck with the derivative at x ** 2 = (2/3) * x. Where did I go wrong?

Edit: 2020-05-08 (Y-m-d):
I did some research and so far it seems I just missed a key principle/concept/rule for solving equalities. Will get back after doing more research. What are your recommendations, advice and thoughts?


Hey, Joachim. Thank you for a detailed question and thank you for solving it yourself. I suggest that you copy the solution into a reply and mark your post as the solution.

Nowhere, really. You just didn’t go deep enough.


As Bruno advised, I would will answer my own question and make it clear that the x values that make 3x^2 - 2x 0 are 0 and 2/3.

I assume that we need to find the values where the derivative is 0 because:

  1. The derivative describes the slope of an equation’s tangent line.
  2. For a tangent line of an equation at a single point at which we apply the slope equation, a limit’s used to describe TWO points with 0 distance from each other because just saying the slope is 0/0 does not give information we can use.
  3. The derivative is set to zero to get critical points because we need to look at a tangent line with slope 0, meaning that at this particular point there is a temporary flatness or plateau, the sign of a potential change in direction of the line’s curve.
  4. According to the graphs in lesson 1 of this mission:
    • the x value in the derivative refers to the value of input x to f(x) and
    • the y' or f'(x) value of the derivative is the slope of a line tangent to the graph of f(x) at a specific x-coordinate. Thus, the x “variable” in the derivative is also the input value to f(x) where the tangent line touches, i.e. the x-coordinate of the point of f(x) that the tangent line touches.

For me, the easy way to find the numbers is by factoring and using the principle that if a*b = 0 then
a = 0 | b = 0 | (a = 0 & b = 0). I got my explanation, which I will write down for you guys to check my understanding, from the links below.
My preferred source:

Solving for the critical points:

Solving the zero values:

To use the zero-factor principle image
Simply factor out the term common to 3x^2 - 2x.
Here x * (3x - 2x) can be fitted to the zero-factor principle as a = x, b = (3x - 2x).
There x = 0 automatically and the solution to make 3x - 2x 0 is x= 2/3.

That’s it for 2020-05-09. Will edit if I find out more or ask a new question if needed.


How did you know that you needed to solve with a quadractic equation? I cannot find that bit of information mentioned anywhere in the course.



We don’t teach everything, some things are prerequisites.

You basically encounter this problem enough times that you’ll just know.

@Bruno That’s not what the marketing on the website states:

Made for: Beginners
Prerequisites: None

I signed up for Dataquest because it promised to teach everything from scratch. The Calculus course has been a struggle for me because it assumes a level of maths knowledge I don’t yet have. Perhaps you could add more content to make it easier for those without a maths background or at least outline the level of maths required beforehand.




Thank you for surfacing this, Jonny.

There have to be some prerequisites: knowing English, knowing how to interact with a computer, and so on. “None” can’t possibly be literally true. We assume some familiarity with mathematics (we don’t teach basic arithmetic for instance).

I will bring this to the team for revision.


Has there been any update in this regard?

Some expansion on when and how to use quadratic equations or some links to external resources would be much appreciated. While working through this mission I have found myself needing to fill in the blanks externally which is a little frustrating when paying for a course that markets itself as something that teaching Data Science from the ground up.

I’m unsure if I should continue if there are going to be more gaps in explanation as the topics increase in complexity.


For what it’s worth: to all people on this page that are struggeling: I totally feel you…

I had 6 years of math in high school (at the highest level in the Dutch educational system) and about three courses of math at university.

But that was 20-25 years ago, and a lot of it I have never used since.

So I’m also struggeling.

I can’t imagine doing this if you have never encountered much math before…

@JoachimRives: thank you for you explanation: it helped me understand the solution.

And thank you for the link to symbolab: I had no idea something like that existed! Very useful!

Some tips: I have found some useful video’s about calculus (among other things) at Khan academy.

And to those of you speaking Dutch: I found some of these Math with Menno video’s helpful as well.


Echoing the other feedback here. I understand there has to be some prerequisites, buth understanding English, knowing how to interact with computers and basic arithmetic is not equivalent to differential calculus. And for for us who either took math a long time ago, or who never took this level of math, some pointers would be very helpful.

Take the statistics and probability courses for instance which are quite detailed and thorough and where external resources are not essential.

Like @mephianna I also recommend Khan Academy. Without them I would not be able to complete this section.

Hey Bruno,

I think that most explanations are very good and detailed. Based on the principle start basic and increase complexity. The complexity generally increases linearly which is exactly what someone who learns need.

In this case it jumps complexity, if you like, not in a linear fashion. You explain basic in a very detailed and clear manner and then the challenge “shifts paradigm” and there is no clear linkage. There was similar situation with Probability - the lottery challenge. The explanation was muddled with a lot of ambiguity and was not clearly built on what was explained prior. I had to look up Khan Academy combinatorics which actually breaks it down quite nicely - exactly in linear fashion.

I know that some of the more challenging topics (like the one explained above with non-linear jump in understanding) are much harder to explain but I would suggest it is just as important as the ones that are easier to explain.

Overall DQ does a great job at teaching DataScience which by the way I publicly acknowledged several times (e.g. review for the course).

At the same time listening to what some individuals suggest in more receptive and understanding fashion might go a long way.

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Hey, Maksym. Thank you for your feedback.

I hear you. This is some of our oldest content and it isn’t up to par with our current standards.
We’re working on a new mathematics course. I don’t know when it will be ready, but it is coming.

Thank you for the kind words.

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For those of you that are struggling with this one, the key information that you need to drag out of the depths of your high school memories is that there is a “Quadratic Formula” that allows you to solve for the two values of x. You may need to google if the name doesn’t jog you memory (I had to). This one was quite a leap but struggle through and the effort will help cement your knowledge. Don’t give up :smile:

Also spotting a quadratic is basically anything where you can make 2 the highest power of x and I agree with @Bruno in that you will pick it up with time and practice.

1 = 3(x^2) is 0 = 3x^2 - 1
x(x - 1) = 4 is 0 = x^2 - x -4

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\begin{eqnarray} 3x^2 - 1 &=& 0 & \\ 3\left(x^2 - \frac{1}{3}\right) &=& 0 &, \text{factor } 3 \\ 3\left(x^2 - \left(\sqrt{\frac{1}{3}}\right)^2\right) &=& 0 &, \text{rewrite } \frac{1}{3} \text{in form of power of 2 in order equation to be in form of } a^2 - b^2 \\ 3(x - \sqrt{\frac{1}{3}})(x + \sqrt{\frac{1}{3}}) &=& 0 &, \text{In form of } a^2 - b^2 = (a - b)(a + b) \\ x - \sqrt{\frac{1}{3}} = 0 &\text{or}& x + \sqrt{\frac{1}{3}} = 0 \\ x = \sqrt{\frac{1}{3}} &\text{or}& x = - \sqrt{\frac{1}{3}}\\ x = \frac{\sqrt{3}}{3} &\text{or}& x = - \frac{\sqrt{3}}{3} \\ \end{eqnarray}
\begin{eqnarray} x^ 2 - x - 4 &=& 0 & \\ (x^2 - x) - 4 &=& 0 \\ (x^2 - 2(1)(\frac{1}{2})x + (\frac{1}{2})^2 )) - (\frac{1}{2})^2 - 4 &=& 0 &, \text{complete the square, in the form of } (x - b)^2 = x^2 - 2(1)(b)x + b^2 = x^2 - 2bx + b^2\\ (x - \frac{1}{2})^2 - \frac{17}{4} &=& 0 \\ (x - \frac{1}{2})^2 - \sqrt{\frac{17}{4}}^2 &=& 0 &, \text{in the form of } a^2 - b^2 \\ \left((x - \frac{1}{2}) - \sqrt{\frac{17}{4}}\right)\left((x - \frac{1}{2}) + \sqrt{\frac{17}{4}}\right) &=& 0 \\ x = \frac{9}{2} + \sqrt{\frac{1}{4}} &\text{or}& x = - \frac{7}{2} - \sqrt{\frac{1}{4}} \\ x = \frac{9}{2} + \frac{\sqrt{4}}{4} &\text{or}& x = - \frac{7}{2} - \frac{\sqrt{4}}{4} \end{eqnarray}
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