I have solved the equation for “3x^2 -2x” and understand the extreme points are [0,2/3]. What I do not understand is how to use the sign chart to identify which value is a relative minimum or maximum. The slopes of values “around” these x values are positive so I am finding it difficult to match the definition of a critical point whose slope changes from positive to negative or vice versa. I could really benefit from understanding how the sign chart is used here.

When we take the derivative of a polynomial function, we get back a function that can tell us the value of the slope of the tangent at any given x in our original function. This is why we set the derivative to zero and solve for x: we are finding the x values where the slope of tangent is equal to zero…which is by definition a critical point!

However, we don’t know if these critical points are relative minimums or relative maximums without looking at the sign of the slope just to the left and right of the critical point. If the slope goes from negative to zero to positive (this is what the sign chart shows us), it means the shape of the graph looks like this: ∪ (IOW, a relative minimum). Whereas, if the sign of the slopes goes from positive to zero to negative, it means the shape of the graph looks like this: ∩ (IOW, a relative maximum).

For x = 0, if we “look a little left”, we see that the slope of the tangent is negative. At x = 0, the slope is 0 (as it should be…it’s a critical point!). And if we “look a little right”, we see the slope of the tangent is negative. Therefore, x = 0 is relative maximum.

Admittedly, this is a lot easier to understand with diagrams. Let me know if you’d like to me to figure out how to slap one together if the above explanation doesn’t help you.

First, thank you for the quick step-by-step response. In defeat, I ended up graphing the function which made it a lot easier to understand why the critical points are relative minimum/maximum.

However, I am still having trouble using the sign graph to show the change in positive and negative. The link to the webpage that demonstrates the sign graph makes sense, but the functions they are solving seem more straight-forward.

Where I start to lose my way: when using the sign chart, I should be solving for the function x^3-x^2 and not the derivative (3x^2-2x)? I found that by plugging x values into the function I was able to determine the y values as negative until x=1. This aligns with the answer in the mission but it isn’t using the slope to determine positive or negative changes.

Is this another way of achieving the same result as measuring changes in slope around the critical values?

No, not defeat…I call that smart! Great job! Getting a visual is super helpful.

I would be a little cautious about following the examples on this site too closely since their aim is to teach how to graph inequalities, not finding the critical points of a polynomial. Ultimately, its value to us is showing what a sign chart is and not so much how to apply it to our use case. You’ll notice there is no mention of slope, critical points, or relative min/max on that page. I would only use that link to understand how to make a sign chart then go back to DQ.

In order to build your sign chart, you should be using the derivative function: 3x^2 - 2x and values of x to the left and right of your critical points (ie the values for x which make the derivative collapse to 0). Once built, your sign chart shows you where your original function (x^3 - x^2) has tangent lines with slopes that are either pos, neg, or zero. Knowing if the slope of the tangent is pos or neg to the left/right of our critical points tells us if that critical point is a relative min or max or neither.

The key point to keep straight in your head while working with these two functions is that plugging in values for x into the original function (x^3 - x^2) will give you y values which then gives you a coordinate pair (x, y) that helps to generate a graph of the original function. Plugging in values for x into the derivative function (3x^2 - 2x) will give you the value of the slope of the tangent at that point in the original function.

For example, if we use x = 1 in the derivative function, we get a result of 1. This means that if you were to draw a tangent line to the graph x^3 - x^2 at x = 1, the slope of that tangent line would be 1 (ie a line at 45 degree angle). Similarly, at x = -1, the slope of the tangent to the graph x^3 - x^2 would be 5 (ie a line at ~78.7 degree angle).

While this is mostly true, this does not tell us anything about our critical points at x=0 and x=2/3. It only tells us that the graph of the original function (x^3 - x^2) will be below the x-axis for all values less than x = 1. And technically speaking, this isn’t true because at x=0 → y=0 and 0 is not less than zero!

For our purposes, the sign chart does not need to be overly complicated or detailed. Let me show you an example:

Find the relative min/max points for the function below:
ƒ(𝑥) = ⅓ 𝑥^{3} + 2𝑥^{2}

| . . . --------[−2]--------[−1]--------[ 0 ]--------[+1]-------- . . . | (values for 𝑥 in the derivative)
| . . . + + + + 0 − − − − − − − − − 0 + + + + + + + + +. . . | (sign of the slope of the tangent)

Since the slope of the tangent went from positive to zero to negative about the point 𝑥 = −2, we know that it is a relative/local maximum. Conversely, since the slope went from negative to zero to positive about the point 𝑥 = 0 we know this is a relative/local minimum.

If you look at screen #4 of this lesson you can see that the graph on the left (𝑥^{3}) does not have a relative min or max since the slope of the tangent goes from pos to zero to pos. Whereas the graph on the right (looks like it might be 2½ 𝑥^{2}) has a relative min at 𝑥=0 because the slope of the tangents goes from neg to zero to pos (going left to right).

This was super helpful I ended up doing a similar exercise for that last section of the mission where I made columns for the x, y, and m values at each point and used the concept of the sign graph moving from left to right to see the change from positive slope, to 0 to negative, and from negative to 0 to positive. Thank you so much for the thorough explanation and examples above they were very helpful!