Derivative of cubic

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Sophiesmark

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Nov 4, 2005
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I'm getting stuck when it comes to finding the derivative of cubic functions.
Like:
f(x)=(x-1)(x-3)(x-4)
for f'(x) I got, (x-3)(x-4)+ (x-1)(x-4)+(x-1)(x-3)
but Im lost as to how I find the second derivative and Im not even totally sure I got the first right.
Help! Thanks :)
 
You've done the first derivative correctly, but....

Have you considered multiplying things out, to get a nicer polynomial form? Putting in a little effort at the beginning could save you a lot of work at the back end. :wink:

Eliz.
 
Is this right cuz I want to get excited or at least feel im going somewhere with all this homework
 
I agree with the idea, but disagree with Sophie's derivitives. I end up with
6x-16
 
I found it!! I picked up something I wasnt supposed to.

It wants me to find the local max and mins and the inflection point. Are the max and mins when the 2nd derivative is zero??
 
Nope. First derivitive = 0 for max & mins.
Second = 0 for inflections.
 
'Fraid not. An error in the constant of f'(x)?
If you look at your original equation the max and mins have to be between the zeros so they are between 1 and 4. More specifically one is between 1 & 3 and one is between 3 & 4.
 
thats right , i forgot....does that mean then that it has no max cuz its a parabola and that the min and the inflection point are the same?
 
Arghhh indeed :!:
Are you picking on me? A parabola can touch/cross the x axis zero, one or two times.
If it crosses zero times, no real max or min.
If it touches one times, one duplicated max or min.
If it crosses two times, one max and one min.
If you have corrected the error in the constant of f'(x) {???} it will cross two times.
You got f''(x) right. It crosses the xaxis at x = 16/6, the inflection point.
If you are still wandering in the wilderness, show what you have for f'(x).
 
Gene said:
Arghhh indeed :!:
Are you picking on me? A parabola can touch/cross the x axis zero, one or two times.
If it crosses zero times, no real max or min.
If it touches one times, one duplicated max or min.
If it crosses two times, one max and one min.
If you have corrected the error in the constant of f'(x) {???} it will cross two times.
You got f''(x) right. It crosses the xaxis at x = 16/6, the inflection point.
If you are still wandering in the wilderness, show what you have for f'(x).
Click to expand...

LOL, I'm not picking on you, I'm getting mad with myself for not comprehending this!!
K, so f'(x)= 3x^2-16x+15, I think I just dont know how to figure out the min and max cuz the three ways I tried gave me values outside the range.
 
As advertised, the constant of f'(x) is NOT 15. I should have asked what you are using for f(x) too. You can do a sorta check on your original multiplication by trying a few x='s like -1,0,2 in both the original ()()() and what you got for x^3+... If any are different answers, you erred. Try again. (If they are all the same it could still be wrong but it is probably right.)
I would like to give you the satisfaction of getting it all for yourself, unless you say "I give up." but little kids will point at you on the streets and laugh if you do.
 
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