1. ## Question about differention, function. Set of values of x.

Hi to all.

Okay, this is a small and simple question, but I just need some help understanding it.

Q) The equation of a curve is y=6x^2-x^3. Find the coordinates of the 2 stationary points on the curve, and determine the nature of these stationary points.

This was part 1 and solved it, getting points (0,0) min and (4,32) max.

However, in part 2, it tells us to state the values of x for which 6x^2-x^3 is a decreasing function of x. Do I have to use the x values found in part 1? And if so, how do I structure the answer?

2. Originally Posted by SolCon
Hi to all.

Okay, this is a small and simple question, but I just need some help understanding it.

Q) The equation of a curve is y=6x^2-x^3. Find the coordinates of the 2 stationary points on the curve, and determine the nature of these stationary points.

This was part 1 and solved it, getting points (0,0) min and (4,32) max.

However, in part 2, it tells us to state the values of x for which 6x^2-x^3 is a decreasing function of x. Do I have to use the x values found in part 1? And if so, how do I structure the answer?

Create three intervals from the x values you found in part 1: $(-\infty,0)$, $(0,4)$, and $(4,\infty)$.

Next, take a value from each of those intervals and plug it into $y^{\prime}$. If $y^{\prime}>0$ at that value, then its increasing on that interval. If $y^{\prime}<0$ at that value, then its decreasing on that interval.

Does this make sense?

But I'm having some trouble understanding this.

When you say y', you mean dy/dx should be applied to the equation, right? And we have each of these values act as subjects? So, for the value of 4, it would like this:

4=12x-3x^2 ?

Also, how would we apply the above (if correct) to infinity and -ve infinity?

4. Originally Posted by SolCon

But I'm having some trouble understanding this.

When you say y', you mean dy/dx should be applied to the equation, right? And we have each of these values act as subjects? So, for the value of 4, it would like this:

4=12x-3x^2 ? no ... you do not set the derivative equal to the x-value where y' = 0. you evaluate the derivative at some value of x in each iinterval just to see if it yields a positive or negative value. for instance, in the interval -infinity to 0, let x = -1

y'(-1) = 12(-1) - 3(-1)^2 = -15 < 0 ... therefore, the function y is decreasing for all values of x in that interval.

...

I understand what you have done. But the book answer is some what confusing. The answer to this question about the set of values of x, is x<0, x>4. In the question, it said decreasing function but x>4 is not so. Also, you have said that we must use y'>0 or y'<0, but the answer being as x> or x< is what's confusing me. Is there something wrong with the book answer?

6. Originally Posted by SolCon

I understand what you have done. But the book answer is some what confusing. The answer to this question about the set of values of x, is x<0, x>4. In the question, it said decreasing function but x>4 is not so. Also, you have said that we must use y'>0 or y'<0, but the answer being as x> or x< is what's confusing me. Is there something wrong with the book answer?
look at the attached graph of the function $y=6x^2-x^3$.

note that for values of $x < 0$ and $x > 4$ the function values are decreasing.

7. Apologies for the late response -_- . Had some things to take care of that were getting in the way.

Any who, the graph is really helpful and now things are more clear. Just a few things though:

1) Do we draw the graph of 6x^2-x^3 or do we use the y' method to find what values are > or < 0?

2) If we draw the graph, how do we know which x-values to take to plug them in the y=6x^2-x^3 equation and tabulate the coordinates? We haven't been given any restriction to work with. In this question, the decrease starts happening before 0. And after 4 , i.e, at x=5. What if the graph were to be such that the decreasing value was to come much later? We'd have to then constantly use the next x value to see when the graph would begin decreasing. What would we have done here if the question were to ask the set of values for an increasing function?

3) In the y' method, for the infinity, we have to keep taking values till we get the y'<0? Do we do this on a Trial and error basis? And do we only take the values that are in infinity? Like for the first interval take values less than 0, and for the third interval, take values greater than 4? Do we at all use 4 or 0 themselves (which will both give us 0 btw) ?

If you could just answer these 3 (hopefully ) last questions, I think this problem will be solved.

Thanks for the help.

8. Originally Posted by SolCon
Apologies for the late response -_- . Had some things to take care of that were getting in the way.

Any who, the graph is really helpful and now things are more clear. Just a few things though:

1) Do we draw the graph of 6x^2-x^3 or do we use the y' method to find what values are > or < 0?
In most beginner calculus course, the latter method is preferred. I'm not saying graphing it and doing things that way is wrong -- but finding where $y^{\prime}$ (or $f^{\prime}(x)$ -- I prefer to use this notation so its less confusing) is > or < 0 is easier to do.

2) If we draw the graph, how do we know which x-values to take to plug them in the y=6x^2-x^3 equation and tabulate the coordinates? We haven't been given any restriction to work with. In this question, the decrease starts happening before 0. And after 4 , i.e, at x=5. What if the graph were to be such that the decreasing value was to come much later? We'd have to then constantly use the next x value to see when the graph would begin decreasing. What would we have done here if the question were to ask the set of values for an increasing function?
If you draw the graph and do it that way, this is what I would recommend (I recommend this to most of my algebra students that I tutor): imagine you have tangent lines to the graph. Now with a pencil, trace along the function and analyze the slope of your pencil. If it looks something like this: \ , then you have negative slope (and thus your function is decreasing as long as your pencil has negative slope). If the pencil looks something like this: / , then you have positive slope (and thus your function is increasing as long as your pencil has positive slope). This way, you have a visual for noticing when your function is increasing and decreasing. Now note that if you have a smooth function, then the pencil will be horizontal when the function goes from increasing to decreasing and vice versa (i.e. at a maximum or minimum value).

3) In the y' method, for the infinity, we have to keep taking values till we get the y'<0? Do we do this on a Trial and error basis? And do we only take the values that are in infinity? Like for the first interval take values less than 0, and for the third interval, take values greater than 4? Do we at all use 4 or 0 themselves (which will both give us 0 btw) ?

If you could just answer these 3 (hopefully ) last questions, I think this problem will be solved.

Thanks for the help.
For your specific problem, you shouldn't use 0 or 4 because we already know that $f^{\prime}(x)=0$ at those values. So what you need to do is pick one point from each interval (and just one). This value will reveal to you the behaviour of the function over the entire interval. So for instance, you can pick $-1\in(-\infty,0)$, $2\in(0,4)$, and $5\in(4,\infty)$ to test (try to go for nice values that yield not-so-messy values if you have to evaluate it by hand; otherwise, you could use a calculator); You could even pick $-\sqrt{2}\in(-\infty,0)$, $\pi\in(0,4)$, $e^3\in(4,\infty)$ and it will still give you the same information on whether or not the function is increasing or decreasing on that interval.

I hope this clarifies things!!

9. Okay, I think that about clears it up.