# Thread: Where is function f(x) = (x^2 - 16)^6 concave up?

1. ## Where is function f(x) = (x^2 - 16)^6 concave up?

The function is ((x^2)-16)^6.

So I did all the relevant calculations and all my others answers about inflexion points, critical points, extreme points, concave down, increasing and decreasing intervals were correct. But my interval for concave up keeps coming up as wrong.

This is the interval I calculated:

(-INF, -4) U (-4, -((4sqrt11)/11) U ((4sqrt11))/11, 4) U (4, INF)

Does anyone know where I have went wrong here? I have been stuck on this problem for a while now.

Thank you!

2. How are you instructed to interpret f(t) = f'(t) = f"(t) = 0? What say you of x = t?

3. Originally Posted by sktsasus
The function is ((x^2)-16)^6.

So I did all the relevant calculations and all my others answers about inflexion points, critical points, extreme points, concave down, increasing and decreasing intervals were correct. But my interval for concave up keeps coming up as wrong.

This is the interval I calculated:

(-INF, -4) U (-4, -((4sqrt11)/11) U ((4sqrt11))/11, 4) U (4, INF)

Does anyone know where I have went wrong here? I have been stuck on this problem for a while now.

Thank you!
There may be slight variations in the definition of "concave up". In this case, the second derivative is zero at -4 and 4, but is positive all around each point. The curve does not actually stop being concave up at those points, so the answer should be (-INF, -((4sqrt11)/11) U ((4sqrt11))/11, INF) .

Please quote the definition you were given for "concave up", as well as the second derivative test for concavity, so we can confirm whether this answer satisfies what you have been taught. (It is possible that there is an inconsistency between your textbook and the answer.)

If it is software that is telling you your answer is wrong, does it also provide any example problems for comparison, or tell you what the correct answer is?

4. Originally Posted by Dr.Peterson
There may be slight variations in the definition of "concave up". In this case, the second derivative is zero at -4 and 4, but is positive all around each point. The curve does not actually stop being concave up at those points, so the answer should be (-INF, -((4sqrt11)/11) U ((4sqrt11))/11, INF) .

Please quote the definition you were given for "concave up", as well as the second derivative test for concavity, so we can confirm whether this answer satisfies what you have been taught. (It is possible that there is an inconsistency between your textbook and the answer.)

If it is software that is telling you your answer is wrong, does it also provide any example problems for comparison, or tell you what the correct answer is?
Thank you! Your answer is correct! I guess I misinterpreted the concavity at -4 and 4.

Yes it was a piece of software on which I was doing this. It tells you when your answer is right or wrong but it does not tell you the correct answer if you are wrong. It does, however, give unlimited attempts, which I suppose makes up for that.

5. Originally Posted by sktsasus
Thank you! Your answer is correct! I guess I misinterpreted the concavity at -4 and 4.

Yes it was a piece of software on which I was doing this. It tells you when your answer is right or wrong but it does not tell you the correct answer if you are wrong. It does, however, give unlimited attempts, which I suppose makes up for that.
Does it provide help, as those I work with do? An example, or a reference to a textbook, should clarify what they want you to do. If it doesn't do that, then it is not teaching you anything, only testing you.

Be sure to look up what "concave up" means. The second derivative gives a way to determine concavity; but it is properly defined as stated here, and the second derivative test is more subtle than it may seem. The key idea for your problem is that the second derivative being zero at one point in an interval does not disqualify the entire interval from being concave up.

6. Originally Posted by Dr.Peterson
Does it provide help, as those I work with do? An example, or a reference to a textbook, should clarify what they want you to do. If it doesn't do that, then it is not teaching you anything, only testing you.

Be sure to look up what "concave up" means. The second derivative gives a way to determine concavity; but it is properly defined as stated here, and the second derivative test is more subtle than it may seem. The key idea for your problem is that the second derivative being zero at one point in an interval does not disqualify the entire interval from being concave up.
The software you are describing sounds quite advanced. Ours seems to be quite simple in comparison. But we do have a textbook through which we can read up on any course material. So, honestly, its not that big of a deal that the software itself does not provide any guide. Its main purpose is to serve as a medium through which we can do our assigned work. That's it really.

And thank you for clarifying to the whole idea of concave up! It makes sense now.

7. Originally Posted by sktsasus
The software you are describing sounds quite advanced. Ours seems to be quite simple in comparison. But we do have a textbook through which we can read up on any course material. So, honestly, its not that big of a deal that the software itself does not provide any guide. Its main purpose is to serve as a medium through which we can do our assigned work. That's it really.

And thank you for clarifying to the whole idea of concave up! It makes sense now.
I'm referring primarily to MyMathLab, which is tied to a textbook, so it can point you to examples in the book, or to examples written to match the problem you are doing. As long as yours is associated with a textbook, that's good.

But make sure that you find the definition you are being taught for concavity, and that the method they teach you for determining it from the second derivative gives the answer the software expects. If not, there could be an inconsistency; let me know if you find one.

8. Originally Posted by Dr.Peterson
I'm referring primarily to MyMathLab, which is tied to a textbook, so it can point you to examples in the book, or to examples written to match the problem you are doing. As long as yours is associated with a textbook, that's good.

But make sure that you find the definition you are being taught for concavity, and that the method they teach you for determining it from the second derivative gives the answer the software expects. If not, there could be an inconsistency; let me know if you find one.
I see.

Yes the textbook's method is fine (matches the answer expected in software). I just overlooked it because this particular function is very weird and involves some strange numbers so I got confused. No excuses there.

9. Originally Posted by sktsasus
… this particular function is very weird and involves some strange numbers …
This function is a 12th-degree polynomial.

x^12 - 96x^10 + 3840x^8 - 81920x^6 + 983040x^4 - 6291456x^2 + 16777216

Its behavior is fairly basic, although it does produce some very large outputs quickly. (For example, from x=0 through x=4, the function begins at 16,777,216 and then drops all the way down to zero. After that, the function increases forever. At x=5.7, it's well-past 16,777,216. At x=6, it has grown to 64,000,000.)

Were the "strange numbers" you mentioned possibly written using Scientific Notation? A version of that notation appears along the y-axis (below), in a graph of this function. We see the number 16,000,000 displayed as 1.6e+07, which means 1.6 times 10^7.

12-deg.jpg

10. Originally Posted by sktsasus
The function is ((x^2)-16)^6.

So I did all the relevant calculations and all my others answers about inflexion points, critical points, extreme points, concave down, increasing and decreasing intervals were correct. But my interval for concave up keeps coming up as wrong.

This is the interval I calculated:

(-INF, -4) U (-4, -((4sqrt11)/11) U ((4sqrt11))/11, 4) U (4, INF)

Does anyone know where I have went wrong here? I have been stuck on this problem for a while now.

Thank you!
Hi sktsasus,

Note that you can easily get the answer without any calculation.

Think first about the graph of the function g(x)=x^2-16=(x-4)(x+4). You can easily see that:

The graph is a parabola (concave up)
• g(x)=0 for x = -4 and x = 4
• g(x) is negative between -4 and +4, and positive outside this interval
• g(x) has a minimum (negative) at x = 0

Now, your function is f(x) = (g(x))^6. Since you are raising (x) to an even power, the negative part will become positive, and the minimum at x = 0 will become a maximum.

Furthermore, since g(x) = 0 at x = -4 and x = 4, f(x) will be zero at the same points (and no other points). Since f(x) >= 0 for all x, these two points are minima.

This allows you to find the shape of the graph, and that is enough to answer the question.

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