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<4`
`f'(x)<0`.
For
`x>4`
`f'(x)>0`.

`\Downarrow\\`
At
`x=4` there's turning point, which is equal
`|4-4|=|0|=0`.