Stationary points I've a question related to stationary point of inflection. A function 
$$ f(x)=ax^5+bx^3+cx $$ has stationary points at $ (-2, 64), (2,-64)~and~(0,0). $  find the values of a, b and c. 
I found the first derivative and equate it to 0. I'm getting same equation when I equate $f'(x)$ to $0$ for $f(2)$ and $f(-2)$. For $f(0)$ I found $c=0$ that is correct. 
Answer provided tells me $a=3$ and $b=-20$.
Please help how I can find this value of $a$ and b .
Thanks,
Arif

You have found $c=0$, so we won't deal with that part. 
We are told that $f(2)=-64$. From this we conclude that 
$$(a)2^5+(b)2^3=-64.$$
The point $(-2,64)$ gives no additional information.
We have $f'(2)=0$. Since $f'(x)=5ax^4+3bx^2$, we have
$$(5a)2^4 +(3b)2^2=0.$$
We have $2$ linear equations in $2$ unknowns. Solve. The solving is easier if we note that the first equation is equivalent to $4a+b=-8$ and the second is equivalent to $20a+3b=0$. 
