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On what interval is the function f(x)=x^3-4x^2+5x concave upward? I found the 1st and 2nd derivative, but now what?

Mathematics
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\(\ \Huge So far, I have: \) \(\ \Huge f'(x)=3x^2-8x+5, \) \(\ \Huge f"(x)=6x-8. \)
Where the second derivative is positive, then you have concave upward, but I just saw what you determined for the derivatives, and they're a bit off. Maybe we should work on that part first.
Okay, so where did I go wrong?

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Other answers:

f'(x) = 3x^2 - 2x - 1. Not sure where where you got the -8 and the 5. Typos? You did the x^3 term correctly, so I have to believe you are familiar with the procedure. derivative of -x^2 is -2x.
Oh! I'm so sorry but I just realized that the question has two typos, let me fix that... Sorry about that!
Maybe the equation in the problem is mistyped? If not, then you should go with my first derivative. We can talk about how we got this, if need be.
Okay, that's the correct f(x). I was looking at a different equation when typing the function...
You're doing well! Now, just look for those values of x for which the second derivative is positive for concave upward.
How do I do that?
So the first derivative doesn't really have much importance in this question? It's the second derivative that's more important?
A really good way to remember if the second derivative is supposed to be positive or negative for concave up (it's positive you want) is to remember y=x^2 as an example. Point 2: to answer your question, yes, it's the second derivative that determines concavity.
What does the first derivative determine then?
And, how do I determine the values for which this function is positive?
For y=x^2, the second derivative is 2, which is positive, so always concave up. The first derivative does 2 things. One, you have to get that first to get the second derivative, and 2), the first derivative will tell you where the function is increasing and decreasing.
Oh okay, I get the first derivative know. I'm still stuck on finding values for which the second derivative is positive..
now*
For determining where the function is positive, even though this is a cubic, you caught a break, because you can factor out "x" and then you are left with a quadratic for the other factor. If you need more help, I can assist. As for finding values for where the second derivative is positive, set 6x - 8 > 0. 6x > 8. x > 4/3.
Okay, but for this problem I don't need to know where the values of the original function is positive, correct? Also, would that mean that the interval for which this function is concave up are \(\ \Huge (\frac{4}{3}, +\infty) ?\)
yes, concave up in that interval. The left "(" should be a "["
As for whether or not you need to know where the function values are positive, that is needed only if you are told to get them. You don't need to know that to determine concavity.
Okay! I thought it was a ( because the inequality was greater than not greater than or equal to
You are understanding this problem well now it seems.
:)
As for +/- values, if you are graphing, the equation is + for positive x and - for negative x. Because the quadratic factor is always positive.
\(\ \Huge Okay, \mathsf T\mathsf h\mathsf a\mathsf n\mathsf k\mathsf s \mathsf f\mathsf o\mathsf r \mathsf a\mathsf l\mathsf l \mathsf t\mathsf h\mathsf e\mathsf h\mathsf e\mathsf l\mathsf p\mathsf ! \)
You are quite welcome! Nice working with you!
It was nice working with you, as well :)

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