# How do you find all critical point and determine the min, max and inflection given f(x)=x^2-8x-10?

Oct 16, 2016

At $x = 4$ is a minimum only

#### Explanation:

$f \left(x\right) = {x}^{2} - 8 x - 10$
Differentiating with respect to x, $f ' \left(x\right) = 2 x - 8$
Differentiating once more, $f ' ' \left(x\right) = 2$ which is $> 0$ so we are expecting a minimum
When $f ' \left(x\right) = 0 \implies 2 x - 8 = 0 \implies x = 4$
When $x < 4 , f ' \left(x\right) < 0$
and when $x > 4 , f ' \left(x\right) > 0$
So we have a minimum
Check
$f \left(3\right) = 3 \cdot 3 - 8 \cdot 3 - 10 = - 25$
$f \left(5\right) = 5 \cdot 5 - 8 \cdot 5 - 10 = - 25$
and $f \left(4\right) = 4 \cdot 4 - 8 \cdot 4 - 10 = - 26$
So we have a min. only