Find the maximum/minimum

**james_bond** · April 29th 2008, 10:30 AM

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

**o_O** · April 29th 2008, 12:25 PM

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

**icemanfan** · April 29th 2008, 12:29 PM

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.

**JaneBennet** · May 1st 2008, 04:06 PM

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

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

Case 1: $\color{white}.\ .$ $2\leq x\leq3$

Then $x-2\geq0$ and by AM–GM, we have

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

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

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

Case 2: $\color{white}.\ .$ $1\leq x\leq2$

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

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

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

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

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

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

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