Question

PLEASE HURRY!!!!! What is the turning point of the graph of f(x) = |x – 4| ?

Answer 1

$f(x)=|x-4|\\ f'(x)=\text{sgn} (x-4)\\ \text{sgn} (x-4)=0\\ x-4=0\\ x=4$

For $x$ $f'(x)$.
For $x>4$ $f'(x)>0$.
$\Downarrow$
At $x=4$ there's turning point, which is equal $|4-4|=|0|=0$.