Differentiation

[eddykk]

What are the x values of all the critical points of this function?:

 

f(x) = x³-6x²+12x-4

 

on the domain Df = [0, 5], as a set {in curly brackets} of the form:

 

{a, b, ...}

 

for real constants a, b, ...

[ydoaPs]

take the first derivative, set it equal to zero. solve for x.

take the second derivative, set it equal to zero. sove for x.

 

 

btw, we won't do your homework for you.

[matt grime]

btw, we won't do your homework for you.

 

I think you just did.

[ydoaPs]

no, i told him how. it is the same thing his teacher would have told him to do.

[JustStuit]

It's really a simple problem. Are you trying to learn on your own? Or did your class start recently? If not, just don't think too hard.

[matt grime]

no, i told him how.

 

 

for this question there is no difference, surely? so I believe you did just do his homework for him; the question is such that there was nothing else you could have done (apart from ask him what the method is he needs to use without specifying it).

[Dave]

take the second derivative, set it equal to zero. sove for x.

 

Sorry, couldn't help but note this as I was going through. You certainly shouldn't do this, assuming you're trying to find the nature of a critical point. Rather, you get the stability of a steady state by evaluating the second derivative at each point, then look at the sign of the derivative.

[ydoaPs]

Sorry, couldn't help but note this as I was going through. You certainly shouldn't do this, assuming you're trying to find the nature of a critical point. Rather, you get the stability of a steady state by evaluating the second derivative at each point, then look at the sign of the derivative.

it didn't say anything about the nature of the critical points. setting the 1st derivative =0 finds the turning points and setting the second =0 finds the points of inflection.

[Dave]

No it doesn't - at least not all of the time. Take the graph [imath]y=x^4[/imath]. Setting the second derivative equal to zero gives [imath]12x^2 = 0 \Rightarrow x = 0[/imath]. This certainly has zero as a root and zero is a critical point, but zero is not a point of inflexion.

 

Besides this, what exactly is the point anyway? All critical points, including points of inflexion are given by setting the first derivative equal to zero and finding the roots. Classifying those points can be quite tricky for some functions, and in many cases you'll have to consider higher order derivatives, or just evaluate points close to your fixed point. Indeed, for the fixed point of the function above, classifying means looking at points either side of the graph.

[ydoaPs]

i've never used the first derivative to find points of inflection. i was taught to use the second.

[Dave]

Just talked on IRC about this, and for the record, not all points of inflection are critical points. I suggest that we simply talk about critical points since that's what the original poster wanted

[ydoaPs]

meh, it's been a while since i have done that stuff.