Question

can someone please explain to me how to get the derivative of absolute(x-3)?

Answer 1

derivative of a linear function is just the same as slope of the line. However, with the absolute value operator, the slope changes at the vertex of the absolute value "v-shape' on the graph.

Answer 2

For big x, like bigger than 3, x-3 will be positive, so abs(x-3) is equal to x-3, and the slope is 1.

For x on the other end of the scale (neg x and x up to but < 3), x-3 is negative, so abs(x-3) = -(x-3) = -x+3 and slope of that section is -1

Answer 3

I don't know if you can express the derivative as a piece-wise function, like "1" for x>3, -1 for x<3

Answer 4

sure

\[f(x) = |x-4| = \left\{\begin{array}{rcc}
3-x & \text{if} & x <3\\
x-3& \text{if} & x \geq 3
\end{array}
\right.
\]
\[f'(x)= \left\{\begin{array}{rcc}-1 & \text{if} & x <3 \\
1& \text{if} & x>3
\end{array}
\right.
\]

Answer 5

i was trying to show off, but instead made a typo
final line is correct though

Answer 6

Is that an "acceptable" form? I couldn't remember... it's clearly "true" (because it's math!!) but I didn't know if there was a normal "style" or form.

Answer 7

lolz...i thought u were trying to get me to learn it and then solve my question

Answer 8

either way thanks @satellite73 and @JakeV8

Answer 9

I was trying to help you learn it, but after the point I left off, I didn't know the "end".. glad to have a rescue from @satellite73 :)

Answer 10

@satellite73 , the derivative is supposed to be \[(x-3)/\sqrt{(x-3)^2}\] but i am stil not getting it

Answer 11

that expression does = 1 for x>3 and -1 for x<3, but I don't know how you get that as the derivative... that is what I meant, though, about whether you could list the derivative as a piece-wise function or if you had to somehow pull it all together into a single form. What you did is that form, but I don't know how you get it...

It appears you can take a derivative of abs function and "force" it into that form... the argument of the abs on the top, and the sq rt of the square of the arg on the bottom.

I just don't recall seeing that before...

Answer 12

it no problem...i was given the assignment but because am new to calculus, i first go to wolframalpha to know the answer i should be working towards. That's where i got the derivative i wrote above from from but my problem now is actually working it out to get that derivative

Answer 13

that's a ridiculous way to write it

Answer 14

:) Feelin' better now...

Answer 15

it is a line
well two lines
one with slope 1, the other with slope -1
that is all

Answer 16

you cannot take the derivative that way, because absolute value is a piecewise function
you must take the derivative separately, one for \(x>3\) the other for \(x<3\)

Answer 17

i think i am just getting myself confused...that derivative will still be equal to 1 and -1

Answer 18

don't try to do it in one step, it wont work
just like if you try to solve \(|x-3|=2\) you have to solve TWO different equations

Answer 19

yes yes, what @satellite73 is saying... :)

Answer 20

yeah..thanx so much and sorry for messing with your heads with my confusion

Answer 21

those absolute value functions are going to "reflect" around the point where the argument of the absolute value (i.e., the whole value of the "stuff" inside) changes from positive to negative. So the resulting function has symmetry across that line, and slope would also be the opposite (negative) on opposite sides of that symmetry line.

Bottom line, it's two steps