1. ## Find the maximum/minimum

Find the local maximum and local minimum of $\displaystyle f(x)=(x-1)(x-2)(x-3)$ using AM-GM inequality.

2. f(x) tends to negative and positive infinity doesn't it?

3. Originally Posted by o_O
f(x) tends to negative and positive infinity doesn't it?
Perhaps the question should read: find the local minimum and maximum, or find the minimum and maximum on an interval.

4. Let’s say we want to find the maximum and minimum of f(x) in the interval $\displaystyle 1\leq x\leq3$. That would make much more sense, wouldn’t it?

Now, rewrite the function as $\displaystyle \mathrm{f}(x)=(x-1)(x-2)(x-3)=(x-2)^3-(x-2)$.

Case 1:$\displaystyle \color{white}.\ .$$\displaystyle 2\leq x\leq3 Then \displaystyle x-2\geq0 and by AM–GM, we have \displaystyle \color{white}.\quad. \displaystyle \frac{(x-2)^3+\left(\frac{1}{\sqrt{3}}\right)^3+\left(\frac {1}{\sqrt{3}}\right)^3}{3}\ \geq\ \frac{x-2}{3} \displaystyle \Rightarrow\ (x-2)^3-(x-2)\ \geq\ -2\left(\frac{1}{\sqrt{3}}\right)^3=-\,\frac{2\sqrt{3}}{9} equality being attained if and only if \displaystyle x-2=\frac{1}{\sqrt{3}}, i.e. if and only if \displaystyle x=2+\frac{1}{\sqrt{3}} Case 2:\displaystyle \color{white}.\ .$$\displaystyle 1\leq x\leq2$

Now $\displaystyle x-2\leq0$ and so we must use $\displaystyle 2-x\geq0$ in the AM–GM formula.

$\displaystyle \color{white}.\quad.$ $\displaystyle \frac{(2-x)^3+\left(\frac{1}{\sqrt{3}}\right)^3+\left(\frac {1}{\sqrt{3}}\right)^3}{3}\ \geq\ \frac{2-x}{3}$

$\displaystyle \Rightarrow\ (2-x)^3-(2-x)\ \geq\ -2\left(\frac{1}{\sqrt{3}}\right)^3=-\,\frac{2\sqrt{3}}{9}$

$\displaystyle \Rightarrow\ (x-2)^3-(x-2)\ \leq\ \frac{2\sqrt{3}}{9}$

Again equality is attained if and only if $\displaystyle 2-x=\frac{1}{\sqrt{3}}$, i.e. if and only if $\displaystyle x=2-\frac{1}{\sqrt{3}}$

Hence the maximum and minimum of f(x) are $\displaystyle \frac{2\sqrt{3}}{9}$ and $\displaystyle -\,\frac{2\sqrt{3}}{9}$, attained when $\displaystyle x=2-\frac{1}{\sqrt{3}}$ and $\displaystyle x=2+\frac{1}{\sqrt{3}}$ respectively.