Question

Find the points on the curve y = 2x3 + 3x2 − 12x + 3 where the tangent line is horizontal.

Answer 1

the tangent is only horizontal at the critical points of the equation, so

`y=2x^3+3x^2-12x+3\implies \cfrac{dy}{dx}=6x^2+6x-12\implies \cfrac{dy}{dx}=6(x^2+x-2) \\\\\\ \stackrel{\textit{setting the derivative to 0}}{0=6(x^2+x-2)}\implies 0=x^2+x-2 \\\\\\ 0=(x+2)(x-1)\implies \boxed{x= \begin{cases} -2\\ 1 \end{cases}} \\\\[-0.35em] ~\dotfill\\\\ y=2(-2)^3+3(-2)^2-12(-2)+3\implies y=-16+12+24+3\implies \boxed{y=23} \\\\\\ y=2(1)^3+3(1)^2-12(1)+3\implies y=2+3-12+3\implies \boxed{y=-4} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \text{\LARGE (-2~~,~~23)\qquad (1~~,~~-4)}~\hfill`