Question

how do you find the derivative of |x+1|-|x-3|?

Answer 1

it is a piecewise function, work in cases

Answer 2

for example if \(x>3\) then \(|x+1|=x+1\) and \(|x-3|=x-3\) so if \(x>3\) you have
\[|x+1|-|x-3|=x+1-(x-3)=-2\] a constant
the derivative of a constant is zero

Answer 3

@satellite73 thats what I thought, but in for my question, it is asking for f'(2.7) of the piecewise function, and the answer given is 2 o-o

Answer 4

that is because if \(-1<x<3\) you have \(|x+1|=x+1\) and \(|x-3|=3-x\)

Answer 5

subtract and get
\[2x-2\] and the derivative is 2

Answer 6

that is the "work in cases" part

Answer 7

@satellite73 why is |x−3|=3−x?

Answer 8

and if a derivative of a constant is 0, why is f'(2.7) not 0 then?

Answer 9

because \(2x-2\) is not a constant

Answer 10

@satellite73 I am still confused about |x−3|=3−x? :s

Answer 11

what is your definition of \(|x-3|\)?

Answer 12

absolute value of x-3; graph is moved 3 units to the right and range is y>0

Answer 13

???

Answer 14

\[|x-3| = \left\{\begin{array}{rcc}
-(x-3) & \text{if} & x < 3 \\
x-3& \text{if} & x \geq 3
\end{array}
\right.\]

Answer 15

ok so |x−3|=3−x because x<3? i think I'll just go review my absolute values unit. thanks!