calculus
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.
Yes, you computed the derivative correctly (assuming you meant to write $(\sin x)^2$ or $\sin^2 x$, not $\sin x^2$).
You need to find the points where the derivative is zero. First note that since $\sin$ is bounded in absolute value by 1, the denominator in your expression for the derivative is never zero. So, the zeroes of the derivative are precisely the zeroes of the numerator of your expression. You thus have to solve the equation $$ (-2\sin x -\sin^2 x)-\cos^2 x=0; $$ a task that is made tractable upon using the Pythagorean identity $\sin^2 x+\cos^2 x=1$. Utilizing this (and a bit of algebra) produces the equivalent equation $$ -2\sin x-1=0. $$ Can you take it from here?
$$\dfrac{dy}{d\theta}\cdot \frac{d\theta}{dx} = \dfrac{dy}{dx}$$ and we want to know when $\dfrac{dy}{dx} = 0$.
$$\frac{dx}{d\theta} = \frac{\cos3\theta}{r},\quad \frac{dy}{d\theta} = \frac{\sin 3\theta}{r} \implies \dfrac{dy}{dx} = \frac{\sin{3\theta}}{r}\cdot \dfrac{r}{\cos 3\theta}$$
Now solve for when $$\require{cancel} \dfrac{dy}{dx} = \frac{\sin{3\theta}}{\cancel{r}}\cdot \dfrac{\cancel{r}}{\cos 3\theta} = = \tan(3\theta)= 0$$
That's going to happen when $\sin 3\theta = 0$ (and note that when $\sin{3\theta} = 0 \iff 3\theta = k \pi \iff \theta = k\pi/3\;$ ($k$ any integer), we have that $\cos 3\theta = \pm 1\iff 3\theta = k\pi \iff \theta = k\pi/3$.)
The points at which the tangents are horizontal and/or vertical are given in polar coordinates, $(r, \theta)$.
Best Answer
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 $$y=\frac{3}{16}x^3-\frac{9}{4}x+4$$