Given $P(x) = x^4+ax^3+bx^2+cx+d$ such $P'(0)=0.$ If $P(-1)
Given $P(x) = x^4+ax^3+bx^2+cx+d$ such that $x=0$ is the only root of $P'(x) = 0.$
If $P(-1)<P(1)\;,$ Then in interval $\left[-1,1\right],$
$\bf{Options::}$
$(a)\;\; P(-1)$ is the minimum and $P(1)$ is the maximum of $P(x).$
$(b)\;\; P(-1)$ is not minimum and $P(1)$ is the maximum of $P(x).$
$(c)\;\; P(-1)$ is the minimum and $P(1)$ is not the maximum of $P(x).$
$(d)\;\;$ Neither  $P(-1)$ is the minimum nor$P(1)$ is the maximum of $P(x).$
$\bf{My\; Try::}$ Given $P(x) = x^4+ax^3+bx^2+cx+d\;,$ Then $P'(x) = 4x^3+3ax^2+2bx+c$
Now Given $P'(0) = 0\Rightarrow c=0$
Now Given $P(-1)<P(1)\Rightarrow 1-a+b+d<1+a+b+d\Rightarrow a>0$
Now I did not Understand How can I solve it, Help me
thanks
 A: Since $x=0$ is the only root of $P'(x)=0$, we have $P'(x)\lt 0$ for $-1\le x\lt 0$ and $P'(x)\gt 0$ for $0\lt x\le 1$. (note that the coefficient of $x^3$ in $P'(x)$ is $4\gt 0$.) Hence, it follows that $P(x)$ is decreasing for $-1\le x\lt 0$ and is increasing for $0\lt x\le 1$.
From this, you should be able to find a correct option.
A: $P(x)$ is a polynomial of degree $4$, which is obviously even, and the leading coefficient is $1$, which is obviously positive. Thus $P(x)$ approaches $+\infty$ both as $x\to-\infty$ and as $x\to+\infty$. $P'(x)$ is zero only at $x=0$, so $P'(x)<0$ for $x<0$ and $P'(x)>0$ for $x>0$. $P(x)$ decreases from $-1$ to $0$ then increases to $1$. Therefore the basic shape of $P(x)$ is a "U" in the interval $-1\le x\le 1$.

There is a global minimum at $0$, by the first derivative test. The first derivative test also tells us that the endpoints $\pm 1$ are both local maxima, but since $f(-1)<f(1)$ the maximum at $-1$ is only local and the one at $1$ is global in the interval $[-1,1]$.
Therefore the answer is

$(b)\;\; P(-1)$ is not minimum and $P(1)$ is the maximum of $P(x)$ in $[-1,1]$

A: Let for example $a=1$, $b=10$ $d=1$: 

$a=1$, $b=-10$ $d=1$:
 
$a=1$, $b=1$ $d=1$: 

$a=-1$, $b=1$ $d=1$: 

