"Prove if $f(x)=ax^3+bx^2+cx+d$ has two critical numbers, their average is the abscissa at the point of inflection."
I've used the quadratic equation to get the average of the critical numbers from the first derivative, which gives me $x=-2b$, and have set the second derivative to zero and obtained the abscissa for the inflection point as $x=\frac{-b}{3a}$
These are obviously not equal. What am I missing?
On
The question concerning the relation between the critical points (when they are present) for a cubic polynomial $ \ f(x) \ = \ ax^3 + bx^2 + cx + d \ \ $ and its point of inflection has already been resolved by Z Ahmed. It may be of interest to consider why such a simple relation holds true.
Since the constant term $ \ d \ $ only affects the "vertical" position of the graph for $ \ f(x) \ \ , $ it has no bearing on the abscissas of the points concerned. The leading coefficient has the effect of "re-scaling" in the $ \ y-$direction ("vertical stretching/compressing"), so it also has no connection to the $ \ x-$coordinates of those points. So we will discuss the properties of the monic polynomial $ \ ax^3 + bx^2 + cx + d \ \rightarrow \ x^3 \ + \ \frac{b}{a}·x^2 \ + \ \frac{c}{a}·x $ $ = \ x^3 \ + \ \beta·x^2 \ + \ \gamma·x \ \ . $
We can begin with just the cubic function $ \ y \ = \ x^3 \ \ , $ which as an odd-power function has "odd symmetry" about the $ \ y-$ axis. This means that the concavity is in opposite directions on either side of that coordinate axis, making the origin the inflection point of the curve. This is the single critical point for this function with $ \ y' \ = \ y'' \ = \ 0 \ \ $ there.
If we next include the other odd-symmetry term, we have $ \ y \ = \ x^3 \ + \ \gamma·x \ \ , $ for which $ \ y' \ = \ 3x^2 \ + \ \gamma \ \ $ and $ \ y'' \ = \ 6x \ \ . $ Again, the concavity is "upward" for $ \ x > 0 \ $ and "downward" for $ \ x < 0 \ \ . $ For $ \ \gamma > 0 \ \ , $ there are no critical points $ \ (3x^2 \ = \ -\gamma \ < \ 0 ) \ \ , $ so we only have the point of inflection at the origin. On the other hand, for $ \ \gamma < 0 \ \ , $ the curve for the function gains two critical points given by $ \ x^2 \ = \ -\frac{\gamma}{3} \ \ $ (with the local maximum at $ \ x = -\sqrt{\frac{-\gamma}{3}} \ $ and the local minimum at $ \ x = +\sqrt{\frac{-\gamma}{3}} \ \ . $ Plainly, the inflection point is midway between the local extrema.
We now "break" the symmetry of the curve about the $ \ y-$axis by inserting the even-symmetry term $ \ \beta·x^2 \ $ into the expression for our function to produce $ \ x^3 \ + \ \beta·x^2 \ + \ \gamma·x \ \ . $ By "completing-the-cube", however, we see that $$ x^3 \ + \ \beta·x^2 \ + \ \gamma·x $$ $$ = \ \ \left[ \ x^3 \ + \ 3·\frac{\beta}{3}·x^2 \ + \ 3·\left(\frac{\beta}{3} \right)^2·x \ + \ \left(\frac{\beta}{3} \right)^3 \ \right] \ + \ \gamma·x \ - \ 3·\left(\frac{\beta}{3} \right)^2·x \ - \ \left(\frac{\beta}{3} \right)^3 $$ $$ = \ \ \left( \ x \ + \ \frac{\beta}{3} \ \right)^3 \ + \ \left( \ \gamma - \ \frac{\beta^2}{3} \ \right) ·x \ - \ \frac{\beta^3}{27} \ \ . $$ Upon introducing the variable $ \ \zeta \ = \ x \ + \ \frac{\beta}{3} \ \ , $ this becomes $$ \zeta^3 \ + \ \left( \ \gamma - \ \frac{\beta^2}{3} \ \right) ·\left( \ \zeta \ - \ \frac{\beta}{3} \ \right) \ - \ \frac{\beta^3}{27} \ \ = \ \ \zeta^3 \ + \ \left( \ \gamma - \ \frac{\beta^2}{3} \ \right) · \zeta \ - \ \frac{\beta \ · \gamma}{3} \ + \ \frac{\beta^3}{27} \ - \ \frac{\beta^3}{27} $$ $$ = \ \ \zeta^3 \ + \ \left( \ \gamma - \ \frac{\beta^2}{3} \ \right) · \zeta \ - \ \frac{\beta \ · \gamma}{3} \ \ . $$
So including the $ \ \beta·x^2 \ $ term translates the purely odd-symmetry cubic curve both "vertically" and "horizontally", which does not destroy the symmetry of the local extrema about the point of inflection. (The "vertical range" between the extrema is altered, but again this is immaterial for the property of interest.) The inflection point lies at $ \ x \ = \ -\frac{\beta}{3} \ = \ -\frac{b}{3a} \ \ , $ with the local extrema at $$ \sqrt{\frac{\frac{\beta^2}{3} \ - \ \gamma}{3}} \ \ = \ \ \frac{\sqrt{ \beta^2 \ - \ 3\gamma}}{3} \ \ = \ \ \frac{\sqrt{ \frac{ b^2}{a^2} \ - \ \frac{3c}{a}}}{3} \ \ = \ \ \frac{\sqrt{ b^2 \ - \ 3ac}}{3a} $$ to either side of it.
This can be thought of as the analogue for cubic polynomials of "all parabolas are similar". In fact, we can take the proposition that the absolute extremum of a quadratic function is midway between its zeroes (when they exist) down by one derivative to arrive at the proposition here for a cubic function.
$$f(x)=ax^3+bx^2+cx+d \implies f'(x)=3ax^2+2bx+c$$ For critical points $$f'(x_{1,2})=0 \implies x_1+x_2=\frac{-2b}{3a} \implies \frac{x_1+x_2}{2}=-b/3a$$ For point of inflexion $f''(x)=6ax+2b=0 \implies x_i=-b/(3a).$ Hence, $x_i=\frac{x_1+x_2}{2}$.