critical points

G

Guest

Guest
Hi guys,

I was given this problem:

Consider the function y = - x^3 + 3x
a)find the critical values
b)where do points of inflection occur
c)If y = 0 then x has what approximate value(s)

Here's what I have so far:
a) for the critical points I have +1 and -1
b) im not sure where to go here, can you lead me in the right direction?
c) only thing I can come up with here is 0? does that sound right?
 
a) Are critical values only min and max? Then you have them.

b) Second derivative = 0

c) I do not understand this piece. Since x = 0 is an obvious solution, the other two are easily found exactly. Generally, though, y = 0 should occur between each pair of critical values identified in a) and possibly not too far outside the most extreme critical values identified in a). So, I would expect y = 0 somewhere on (-1,1) (obviosuly, we know x = 0 is there), and on (1,1+(not too much)) and on (-1-(not too much),-1). I'm being deliberately vague, since I'm sure I can design a function that will have y = 0 very, very far to the right of the maximum critical value, but generally, on a textbook problem, you shouldn't have to go hunting very far.
 
Tk,
Do you mean for b) that points of inflection occur when the second derivative is set to zero?
so if f' = 3x^2 - 3
f" = 6x
6x = 0, so x = 0? So 0 would be the only inflection point?

For c) my possible answers are:
a. x = 10
b. x = ±1.732
c. x = 3
d. both a and b above

so from what you said b only applies?
 
Do you mean for b) that points of inflection occur when the second derivative is set to zero?
so if f' = 3x^2 - 3
f" = 6x
6x = 0, so x = 0? So 0 would be the only inflection point?

For c) my possible answers are:
a. x = 10
b. x = ±1.732
c. x = 3
d. both a and b above

so from what you said b only applies?
Click to expand...

Yes and yes. (Hi, tk. I admire your work -- as usual! :D )
 
Hmpf! Then I don't get it, since we know already that y = 0 when x = 0. where it that solution? Maybe it's a typo on the worksheet.
 
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