Stationary Points Possible Values

Jul 2019
5
0
Australia
Hi,

I am having trouble with part d) of this question. It follows on with other parts of a question which I have attached. I have written that 'p' can indeed have stationary points but am not sure what the possible values of 'p' could be. If anyone can list these possible values that would be great, as this is apart of a big assignment that is due soon. Please note this is from question 4 from the first picture.

Thank You and Kind Regards.
 

Attachments

Dec 2014
131
102
USA
To be clear, part (d) asks about the number, p, of possible stationary points of the quartic polynomial function $y=ax^4+bx^3+cx^2 +dx+e$

Stationary points occur where $\dfrac{dy}{dx} = 0$. The derivative of a quartic polynomial function is a cubic polynomial function.

For such a cubic function, what are the possible number of zeros?
 
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Jul 2019
5
0
Australia
Hi,

Thanks for that, I posted the question another palce online and they said that the possible values are P=1,2,3 is this true?
 
Jul 2019
5
0
Australia
Hi,

Would the answer just be P = 1, 2 or 3 given the points we have been given of where it passes
 
Dec 2014
131
102
USA
Hi,

Would the answer just be P = 1, 2 or 3 given the points we have been given of where it passes
Note that part (d) has no relation to parts (a), (b), and (c) and the given points.

The polynomial in parts (a), (b), and (c) is $y=ax^2+bx^3+cx^2+d$. There is no linear term.

The polynomial in part (d) is $y=ax^2+bx^3+cx^2+dx + e$. Not the same as above.
 
Jul 2019
5
0
Australia
Is it okay if you could tell me the answer, I'm in kind of a short space of time to get this done, and it is the only thing I have left.
 

topsquark

Forum Staff
Jan 2006
11,569
3,453
Wellsville, NY
Is it okay if you could tell me the answer
No. I'm sorry, but that's Forum policy.

In post #2 Cervesa asked you how many points you would have for a cubic. Do you know how to find the answer to that question? (Hint: You want to set dy/dx = 0. What kind of polynomial equation do you get for this condition?)

-Dan
 
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