Analysis by Its History_exercise 2.1

(Euler 1755,page261).Study the functions
\begin{equation}
\label{eq:29.14.51}
y=x^4-8x^3+22x^2-24x+12
\end{equation}

\begin{equation}
\label{eq:29.14.52}
y=x^5-5x^4+5x^3+1
\end{equation}

Find maxima,minima,convex downward regions,inflection points.

Solve1:$y'=4x^3-24x^2+44x-24$.$y''=12x^2-48x+44$.When
\begin{equation}
\label{eq:29.15.16}
4x^3-24x^2+44x-24=0
\end{equation}
That is,
\begin{equation}
\label{eq:29.15.18}
4x^3-4x^2-20x^2+20x+24x-24=0
\end{equation}
That is,
\begin{equation}
\label{eq:29.15.19}
(x-1)(4x^2-20x+24)=0
\end{equation}
So $x_1=1,x_2=2,x_3=3$.$y''(x_1)>0,y''(x_2)>0,y''(x_3)>0$,so there is no maxima to the function $y=x^4-8x^3+22x^2-24x+12$,the minima of the function is $1,2,3$.

Let $y''=12x^2-48x+44>0$,so the convex downward regions are $(\frac{6-\sqrt{3}}{3},\frac{6+\sqrt{3}}{3})$.The inflection points are $\frac{6-\sqrt{3}}{3}$ and $\frac{6+\sqrt{3}}{3}$.

Solve2:$y'=5x^4-20x^3+15x^2$.$y''=20x^3-60x^2+30x$.When
\begin{equation}
\label{eq:29.15.48}
y'=5x^4-20x^3+15x^2=0
\end{equation},
$x_1=0,x_2=1,x_3=3$.$y''(x_1)=0,y''(x_2)<0,y''(x_3)>0$,so $x_2=1$ is a maxima,$x_3=3$ is a minima.What about $x_1=0$?It is easy to verify that $x_1=0$ is neither a maxima nor a minima(How?hint:Observe \ref{eq:29.14.52}).

Let $y''=20x^3-60x^2+30x>0$,then $x(2x^2-6x+3)>0$,then $x(x-\frac{3-\sqrt{3}}{2})(x-\frac{3+\sqrt{3}}{2})>0$.So the convex downward regions are $(0,\frac{3-\sqrt{3}}{2})\bigcup (\frac{3+\sqrt{3}}{2},+\infty)$.The inflectino points are $0,\frac{3-\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}$.