Answer to Question #338887 in Calculus for Junior

Question

For what values of 𝑐 does the curve 𝑓(𝑥) = 2𝑥^3 + 𝑐𝑥^2 + 2𝑥 have maximum and minimum points?

Show that the minimum and maximum points of every curve in the family of polynomials

𝑓(𝑥) = 2𝑥^3 + 𝑐𝑥^2 + 2𝑥 lie on the curve 𝑦 = 𝑥 − 𝑥^3.

Answer 1

We point out that the function increases for large values of $|x|$. It is due to the fact that the term $2x^3$ increases as $x$ increases. Consider the derivative: $f'=6x^2+2cx+2$. In case we consider the equation $f'=0$, we get the following discriminant: $D=4c^2-48=4(c^2-12)$. Thus, for $|c|\leq\sqrt{12}$, $f'\geq0$. It means, that for those values of $c$ the function $f(x)$ increases for all values of $x$. Thus, minimum and maximum points are absent. For $|c|>\sqrt{12}$ we get: $x_1=\frac{-2c-\sqrt{D}}{12}$, $x_2=\frac{-2c+\sqrt{D}}{12}$. The function $f$ increases for $x\in(-\infty,x_1)$ and for $(x_2.+\infty)$. It decreases for $x\in(x_1,x_2)$. Thus, $x_1$ is a local maximum and $x_2$ is a local minimum. Consider the points of intersection of the functions $f(x)=2x^3+cx^2+2x$, $y=x-x^3$. We receive: $3x^3+cx^2+x=0$. As we can see, the equation is the same as the equation $f'=0$. It means that minimum and maximum points can be found from intersection of curves $f(x)=2x^3+cx^2+2x$ and $y=x-x^3$.

Answer: In case $|c|\leq\sqrt{12}$, function $f(x)$ has no minimum and maximum points. For $|c|>\sqrt{12}$, $x_1=\frac{-2c-\sqrt{D}}{12}$, where $D=4c^2-48=4(c^2-12)$, is a local maximum and $x_2=\frac{-2c+\sqrt{D}}{12}$ is a local minimum.