In my experience, I have usually dealt with finding or analyzing tangents to a curve. In this question I've encountered, I am asked to analyze a curve given it is a tangent to the line y=x.
Here is the curve:
$y=(x^3/3)+ax+b. $
I am tasked to prove that the above conditions imply that:
$4(a-1)^3+9b^2=0$
I have deduced some things from the implication, but I feel that I have not deduced enough! Here is what I have understood so far:
The slope of $y=x$ is 1. At the point the curve is tangent to this line, it's slope is 1. Therefore, I differentiated the equation of the curve with respect to $x$ and reached the following conclusion:
$x^2+a=1$ so $x=\pm \sqrt(1-a) $
Substituting this result in the equation of the curve, I get
$y=((\pm \sqrt(1-a))^3/3)+a(\pm \sqrt(1-a))+b. $
This is something, but does not seem to get me closer to deducing the required.
At the point where the curve is tangent, we also have
$y=x=\pm \sqrt(1-a)=((\pm \sqrt(1-a))^3/3)+a(\pm \sqrt(1-a))+b. $
My question is thus: Am I not considering an implication of the curve being a tangent to this line that will help me deduce the required?
Suppose the point in question is at $x = p$. Then since the two curves intersect at this point and since the slope of their tangent lines are equal at this point, we have the system: $$ \begin{cases} p = \frac{1}{3}p^3 + ap + b \\ 1 = p^2 + a \end{cases} \qquad\iff\qquad \begin{cases} -b = \frac{1}{3}p^3 + (a - 1)p \\ a - 1 = -p^2 \end{cases} $$ Substituting in the $(a - 1)$, we get: $$ -b = \frac{1}{3}p^3 + (-p^2)p = \frac{-2}{3}p^3 \iff p = \sqrt[3]{\frac{3}{2}b} $$ Substituting this back, we have: $$ a - 1 = -\left( \sqrt[3]{\frac{3}{2}b} \right)^2 \iff (a - 1)^3 = -\frac{9}{4}b^2 \iff 4(a - 1)^3 + 9b^2 = 0 $$ as desired.