# What are the extrema of f(x)=(x - 4)(x - 5) on [4,5]?

Nov 5, 2015

The extremum of the function is (4.5 , -0.25)

#### Explanation:

$f \left(x\right) = \left(x - 4\right) \left(x - 5\right)$ can be rewritten to $f \left(x\right) = {x}^{2} - 5 x - 4 x + 20 = {x}^{2} - 9 x + 20$.

If you derivate the function, you will end up with this:
$f ' \left(x\right) = 2 x - 9$.
If you don't how to derivate functions like these, check the description further down.

You want to know where $f ' \left(x\right) = 0$ , because that's where the gradient = 0.

Put $f ' \left(x\right) = 0$ ;
$2 x - 9 = 0$
$2 x = 9$
$x = 4.5$

Then put this value of x into the original function.
$f \left(4.5\right) = \left(4.5 - 4\right) \left(4.5 - 5\right)$
$f \left(4.5\right) = 0.5 \cdot \left(- 0.5\right)$
$f \left(4.5\right) = - 0.25$

Crach course on how to derivate these types of functions:
Multiply the exponent with the base number, and decrease the exponent by 1.

Example:
$f \left(x\right) = 3 {x}^{3} - 2 {x}^{2} - 2 x + 3$
$f ' \left(x\right) = 3 \cdot 3 {x}^{3 - 1} - 2 \cdot 2 {x}^{2 - 1} - 1 \cdot 2 {x}^{1 - 1}$
$f ' \left(x\right) = 9 {x}^{2} - 2 x - 2 {x}^{0}$
$f ' \left(x\right) = 9 {x}^{2} - 2 x - 2$