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

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The discussion revolves around solving the equation \(\frac{f(-p)+f(p)}{2}=f(0)\) for a cubic function \(f(x) = ax^3 + bx^2 + cx + d\) given three collinear points. The initial attempt incorrectly assumed \(f(-p) + f(p) = 0\). By letting \(p = 1\) and analyzing the function, it is shown that \(b\) must equal zero, simplifying the function to \(f(x) = ax^3 + cx + d\). Consequently, it is established that \(\frac{f(-x)+f(x)}{2}=d=f(0)\), confirming the relationship holds true for the cubic function. The discussion effectively demonstrates the conditions for collinearity and the implications for the coefficients of the cubic function.
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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]
 
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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:

$$\frac{f(-x)+f(x)}{2}=d=f(0)$$
 
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