49 lines
1.3 KiB
TeX
49 lines
1.3 KiB
TeX
\beginsection 18.3
|
|
|
|
Use calculus to find the maximum and minimum of
|
|
$f(x)=x^3-6x^2+9x+1$ on $[0,5)$.
|
|
|
|
\medskip
|
|
|
|
The maximum and minimum occur where $df/dx=0$.
|
|
$$df/dx=3x^2-12x+9=0$$
|
|
This equation has two solutions, $x=1$ and $x=3$.
|
|
That was easy. Now we have to work harder.
|
|
|
|
1. Use the ``first derivative test'' to see if $x=1$ is a minimum or maximum.
|
|
Since $f'(0)>0$ and $f'(2)<0$ we conclude that $x=1$ is a local maximum.
|
|
|
|
2. Repeat for $x=3$.
|
|
Since $f'(2)<0$ and $f'(4)>0$ we conclude that $x=3$ is a local minimum.
|
|
|
|
3. Now compute $f$ at $x=0,1,3,5$ and compare.
|
|
|
|
$f(0)=1$
|
|
|
|
$f(1)=5$
|
|
|
|
$f(3)=1$
|
|
|
|
$f(5)=21$.
|
|
|
|
We observe that $f(5)=21$ is the maximum
|
|
but since the interval is open on the 5 side, we conclude that $f$
|
|
has no maximum.
|
|
The function $f$ has two minima, one at $f(0)$ and the other at $f(1)$.
|
|
|
|
4. I just realized that I could have skipped steps 1 and 2.
|
|
The computation of $f$ at $x=0,1,3,5$ is all that's needed.
|
|
Also, the ``second derivative test'' is an easier way to find out if $x_0$ is a
|
|
local minimum or maximum:
|
|
|
|
$f''(x_0)>0$ means $x_0$ is a local minimum,
|
|
|
|
$f''(x_0)<0$ means $x_0$ is a local maximum.
|
|
|
|
|
|
\bigskip
|
|
BTW, {\tt gnuplot} was a big help in solving the problem.
|
|
|
|
\centerline{\tt plot [0:5] f(x)=x**3-6*x**2+9*x+1, f(x), g(x)=3*x**2-12*x+9, g(x)}
|
|
|