Given this Quartic function $p(x)=ax^4+3x^3+6x+2$ , we need to solve according to the information given about $'a'$ and about the polynomial attribute(sorry if the word is not correct) in the interval $(-\infty,+\infty)$. We need to choose 1 correct answer
A. The polynomial decreases in the interval
B.No local maxima or minima
C.The polynomial increases in the interval
D. there is no global maxima
E. there is no global minima
The information about $a$:
- $'a'$ is negative
- $'a'$ is positive
My try for the first part($a$ is negative):
I checked the limits when ${x \to +\infty }$ and ${x \to -\infty }$ because if we have $(-\infty,+\infty)$ and the limits are $\lim \limits_{x \to +\infty }p(x)=-\infty$ and $\lim \limits_{x \to -\infty }p(x)=-\infty$ then it means there exists a maxima in $(-\infty,+\infty)$ .
$\lim \limits_{x \to +\infty }ax^4+3x^3+6x+2=-\infty$ (since we only deal with the highest power and $a$ is negative
$\lim \limits_{x \to -\infty }ax^4+3x^3+6x+2=-\infty$( for the same reason and here the even power make it positive)
according to this we have a maxima , and since $a$ is negative we know that our function will look similar in a way to a parabola with a frown which is our global maxima. and of course the derivative would be $p'(x)=4ax^3+9x^2+6$ but atleast for me I could not see a way that the derivative would help me.
my conclusions were that there is no local maxima or minima(B) , there is no global minima(E) , there is a global mxima(D) , for A and C I do not know.
for the second part ($a$ is positive):
According to the derivative $p'(x)=4ax^3+9x^2+6$ the function increases for every x.
so I will also check to see if there is a minimum or maximum according to a similar thing but this time if $\lim \limits_{x \to +\infty }p(x)=+\infty$ and $\lim \limits_{x \to -\infty }p(x)=+\infty$ then it means we have a minima but no maxima in the interval.
$\lim \limits_{x \to +\infty }ax^4+3x^3+6x+2=+\infty$ (since again we only deal with the highest power and $a$ is positive this time)
$\lim \limits_{x \to -\infty }ax^4+3x^3+6x+2=+\infty$(for the same reason)
so according to this we have a minima , no maxima , the function increases to every x but I still do not know the correct answers.
Thank you for the help , I would like to know if my ways or at least the thoughts were correct in some way And I would appreciate any tips and advice on how to approach this better and how to solve $A$ and $C$.
Sorry for the English mistakes.
Best Answer
Problem can be attacked by meta-cheating based on the idea that there is only 1 correct answer. Clearly $ax^4$ is the dominant term, so if $a$ is positive, there is no global maxima and if $a$ is negative, there is no global minima.
It remains to
Options A and C are easily dismissed, because for the generic cubic represented by $p'(x)$, how in the world could it be guaranteed that the cubic is always less than zero or always greater than zero.
Option B is similarly dismissed. How in the world could one guarantee that the pertinent cubic has no real root?
From what I can see, your overall analysis seems sound: I saw no obvious mistakes.
Exception
The issue is (for example) whether it can be guaranteed that $p'(x)$ is always positive, regardless of the value of $a$. If $a$ is positive (non-zero), then for large negative $x$, $p'(x)$ will be negative and for large positive $x$, $p'(x)$ will be positive.
By focusing on whether $p'(x)$ is increasing or decreasing, you are mistakenly focusing on whether $p''(x)$ is positive or negative, rather than focusing on whether $p'(x)$ is positive or negative.