simple one variable polynomial function: finding coefficients

Question

Write down the values of the coefficient $b$ and $d$ for which the graph of $f(x)$ with equation $f(x) = ax^3 + bx^2 + cx + d$ would be symmetric about the origin.

Because this is an odd function, it leads me to the bx^2+d=0 equation. I am not sure that I am on the right way to the solution.

Thank you!

Unfortunately, what you want is an odd function, not an even function. — angryavian, May 07 '19 at 19:34
you are right, I just miswrote it, it's an odd function, and I used the -f(x)=f(-x) condition. — TTomi, May 07 '19 at 19:37
-f(x)=f(-x) -ax3-bx2-cx-d=a(-x)3+bx2+c(-x)+d which leads to 0=bx2+d — TTomi, May 07 '19 at 19:43
if x can be any real number, b=0 and d=0 are the only solutions? — TTomi, May 07 '19 at 19:44
@TTomi Yes, since $bx^2+d=0$ must hold for all $x$, we must have $b=d=0$. — angryavian, May 07 '19 at 20:04

score 0 · Answer 1 · answered May 07 '19 at 19:39

0

Yes you are right. Its just that you are using the word even instead of odd.

answered May 07 '19 at 19:39

Akul Singhal

306

Thank you SR810! – TTomi May 07 '19 at 20:54

score 0 · Accepted Answer · answered May 07 '19 at 20:38

0

You want that $f(-x)=-f(x)$ for any $x$. In particular, $f(-1)=-f(1)$, so $$ -a+b-c+d=-a-b-c-d $$ whence $b+d=0$. Similarly, you need $f(-0)=-f(0)$, so $$ d=-d $$ whence $d=0$.

The two conditions imply $b=d=0$. The function $f(x)=ax^3+cx$ is odd for any value of $a$ and $c$, so you're done.

answered May 07 '19 at 20:38

egreg

238,574

Thank you egret!! – TTomi May 07 '19 at 20:55

simple one variable polynomial function: finding coefficients

2 Answers2