# What are the absolute extrema of f(x)=(x-2)(x-5)^3 + 12in[1,4]?

Jun 22, 2016

Local Minima. is $- \frac{2187}{128.}$
Global Minima$= - \frac{2187}{128} \cong - 17.09 .$
Global Maxima $= 64.$

#### Explanation:

For extrema, $f ' \left(x\right) = 0.$

$f ' \left(x\right) = \left(x - 2\right) \cdot 3 {\left(x - 5\right)}^{2} + {\left(x - 5\right)}^{3} \cdot 1 = {\left(x - 5\right)}^{2} \left\{3 x - 6 + x - 5\right] = \left(4 x - 11\right) {\left(x - 5\right)}^{2.}$

$f ' \left(x\right) = 0 \Rightarrow x = 5 \notin \left[1 , 4\right] ,$ so no need for further cosideration & $x = \frac{11}{4.}$

$f ' \left(x\right) = \left(4 x - 11\right) {\left(x - 5\right)}^{2} , \Rightarrow f ' ' \left(x\right) = \left(4 x - 11\right) \cdot 2 \left(x - 5\right) + {\left(x - 5\right)}^{2} \cdot 4 = 2 \left(x - 5\right) \left\{4 x - 11 + 2 x - 10\right\} = 2 \left(x - 5\right) \left(6 x - 21\right) .$

Now, $f ' ' \left(\frac{11}{4}\right) = 2 \left(\frac{11}{4} - 5\right) \left(\frac{33}{2} - 21\right) = 2 \left(- \frac{9}{4}\right) \left(- \frac{9}{2}\right) > 0 ,$ showing that, $f \left(\frac{11}{4}\right) = \left(\frac{11}{4} - 2\right) {\left(\frac{11}{4} - 5\right)}^{3} = \left(\frac{3}{2}\right) {\left(- \frac{9}{4}\right)}^{3} = - \frac{2187}{128} ,$ is Local Minima.

To find Global Values, we need $f \left(1\right) = \left(1 - 2\right) {\left(1 - 5\right)}^{3} = 64 ,$ & $f \left(4\right) = \left(4 - 2\right) {\left(4 - 5\right)}^{3} = - 2.$

Hence, Global Minima $= \min$ { local minima, f(1), f(4)}=min{-2187/128,64, -2}=min{-17.09, 64, -2}=-2187/128~=-17.09

Global Maxima $= \max$ { local maxima ( which doesn't exist), f(1), f(4)}=max{64, -2}=64.