Find the cubic function $y = ax^3 + bx^2 + cx + d$ whose graph has horizontal tangents at the points $(-2,7)$ and $(2,1)$.
I find the derivative and set equation to zero, but then that only gives me one solution. I can't find all solutions for some reason.
First, find the derivative of $y$ with respect to $x$: $$y'=3ax^2+2bx+c$$ And because of the tangent points given, we know that $$3a(-2)^2+2b(-2)+c=0$$ $$3a(2)^2+2b(2)+c=0$$ And since the graph of the cubic must pass through those points, $$a(-2)^3+b(-2)^2+c(-2)+d=7$$ $$a(2)^3+b(2)^2+c(2)+d=1$$ When we simplify these equations, we get as system of four equations with four variables to solve for: $$12a-4b+c=0$$ $$12a+4b+c=0$$ $$-8a+4b-2c+d=7$$ $$8a+4b+2c+d=1$$ This system can be solved easily using matrices. After solving, we get $$a=\frac{3}{16}$$ $$b=0$$ $$c=-\frac{9}{4}$$ $$d=4$$ So the cubic must be $$y=\frac{3}{16}x^3-\frac{9}{4}x+4$$
