Finding the local and absolute values of a polynomial on a closed finite interval Problem:
Find the local and absolute extreme values of the $y$ on the interval $[0,5]$.
$$ y = x^3 - 9x^2 + 24x - 2 $$
Answer:
To see a plot of this function, click on the link below.
enter link description here
Based upon the plot, I claim that $x = 0$ is a local minimum and $x = 5$ is a local maximum. Now, I am going to look
for another local minimum or another local maximum.
\begin{align*}
y' &= 3x^2 - 18x + 24 \\
y' &= 0 \\
3x^2 - 18x + 24 &= 0 \\
x^2 - 6x + 8 &= 0 \\
(x-4)(x-2) &= 0 \\
\end{align*}
Hence, I claim that $x = 2$ is a local maximum and $x = 4$ is a local maximum. Now, I find $y(2)$.
\begin{align*}
y(2) &= 2^3 - 9(4) + 24(2) - 2 \\
y(2) &= 8 - 36 + 48 - 2 \\
y(2) &= 18 \\
y(5) &= 5^3 - 9(25) + 24(5) - 2 \\
y(5) &= 125 - 225 + 24(5) - 2 \\
y(5) &= 18
\end{align*}
Hence the absolute extreme values of the function on the interval is $x = 2$ and $x = 5$.
Do I have this right?

For $x\in(0,5)$ we have
$$y' = 3x^2 - 18x + 24=0 \implies x_1=2\:,x_2=4 \implies f(x_1)=18\:, f(x_2)=14$$
and since
$$y'' = 6x - 18\implies  f''(x_1)<0\:, f''(x_2)>0$$
then on the boundary
$$x_0=0\:,x_3=5 \implies f(x_0)=-2\:, f(x_3)=18$$
Therefore

*

*$x_1$ is a point of local maximum


*$x_2$ is a point of local minimum


*$x_0$ is a point of absolute minimum $(-2)$


*$x_1$ and $x_3$ are point of absolute maximum $(18)$
