Solve for a,b,c,d Solve \frac{f(-1)+f(1)}{2}=f(0) for a, b, c, d

[Bushy]

For a cubic function \(\displaystyle y=f(x)\) three points A, B and C lie on a straight line with respective coordinates (-p,f(p)),(0,f(0)) and (p,f(p)) where p is a non zero constant.

a. Show that \(\displaystyle \frac{f(-p)+f(p)}{2} =f(0)\)

I tried

\(\displaystyle \frac{f(-p)+f(p)}{2} =\frac{0}{2} = 0 = f(0)\)

That doesn't seem right

b. Let p =1 and \(\displaystyle f(x) = ax^3+bx^2+cx+d \)

i. Show that b=0

ii. hence \(\displaystyle \frac{f(-x)+f(x)}{2}=f(0)\)

 

[MarkFL]

I think you meant to give the coordinates of $A,\,B,\,C$ as $(-p,f(-p)),\,(0,f(0)),\,(p,f(p))$. If the 3 points are collinear, then we can write:

\(\displaystyle f(\pm p)=f(0)\pm k\)

Let's let:

\(\displaystyle f(x)=ax^3+bx^2+cx+d\)

So, we find that:

\(\displaystyle f(0)=d\)

And then we must have:

\(\displaystyle ap^3+bp^2+cp=k\)

\(\displaystyle -ap^3+bp^2-cp=-k\)

Adding these two equations, we find:

\(\displaystyle b=0\)

And so:

\(\displaystyle f(x)=ax^3+cx+d\)

Then we find:

\(\displaystyle \frac{f(-x)+f(x)}{2}=d=f(0)\)