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 $$y=f(x)$$ 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 $$\frac{f(-p)+f(p)}{2} =f(0)$$

I tried

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

That doesn't seem right

b. Let p =1 and [math]f(x) = ax^3+bx^2+cx+d [/math]

i. Show that b=0

ii. hence [math]\frac{f(-x)+f(x)}{2}=f(0)[/math]

 

[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:

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

Let's let:

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

So, we find that:

$$f(0)=d$$

And then we must have:

$$ap^3+bp^2+cp=k$$

$$-ap^3+bp^2-cp=-k$$

Adding these two equations, we find:

$$b=0$$

And so:

$$f(x)=ax^3+cx+d$$

Then we find:

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