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

1 answer:

8 0

$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 $|4-4|=|0|=0$ .

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