Question

find the derivative of f(x)= |x-3| +2
help plsss :D

Answer 1

depends on the domain, since the absolute value is a piecewise function

Answer 2

if \(x>3\) then \(|x-3|=x-3\)

Answer 3

whereas if \(x<3\) then \(|x-3|=3-x\)

Answer 4

so your derivative is either 1 or -1 depending on whether \(x>3\) or \(x<3\)

Answer 5

\(\large\color{black}{ \displaystyle \frac{d}{du}\left[~|u|~\right]=\frac{u}{|u|} \cdot u' }\)

Answer 6

you can prove that, by using the definition \(\large\color{black}{ \displaystyle |u|=\sqrt{u^2}}\)

Answer 7

alright that makes sense, do you guys know the mean value theorem

Answer 8

heard of it yes

Answer 9

im given the interval [1,6] and i need to figure out which hypothesis is not met

Answer 10

is it that f'(x) is not differentiable?

Answer 11

ok that is a reasonable question
we have to check that the function is continuous on the closed interval and differentiable on the open one

Answer 12

don't be confused by that
differentiable doesn't mean "i can find the derivative" differentiable means "the derivative exists on the entire interval"

Answer 13

what is the function?

Answer 14

Given a function F(x)

in some interval [a,b], a<=c<=b
there exists a value c, such that F(c)= (F(b)+F(a))/2

Answer 15

yea but it doesnt right? cuz we just said its either 1 or -1?

Answer 16

oooh it was the function above!

Answer 17

yes, the answer is NO because your function is not differentiable at \(x=3\)

Answer 18

alright thanks for clearing things up :D

Answer 19

yw

Answer 20

@dan815 over two?

Answer 21

yeah its the average value

Answer 22

oh no that is not what the mvt says at all

Answer 23

isnt it f(b)-f(a)/b-a = f'(c) for the mean value theorem?

Answer 24

in fact, not to be argumentative, but it is not even close

Answer 25

@Noodles24 that is the CONCLUSION of the mvt

Answer 26

oh okay srym b-a i see

Answer 27

yea i thought thats what he was trying to say

Answer 28

if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\) then
there is a \(c\in (a,b)\) with
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]

Answer 29

i misunderstood what MVT was for again hehe,
So i guess this means we are saying there is a F(c) such that it is the average change of the function on the interval [a,b] I previously thought MVT is just the average value not the average change xD

Answer 30

i got another quick question for ya

Answer 31

if the derivation of something is just like one number, such as 1, is it differentiable?

Answer 32

idk if u would ever have that though

Answer 33

yeah if its non differentiable, we can have like an infinity or some un accounted change of the average value showing up

Answer 34

alright i got another kinda stupid question

Answer 35

cos(x)-xsin(x)=0

Answer 36

how u solve for x lol, i feel like this is something easy but i cant seem to rememeber

Answer 37

x=cos(x)/sin(x)
1/x=tan(x)

so 1 over the angle = opp/adj

Answer 38

or adj/opp = x

Answer 39

i understand but i need like a numerical value for x

Answer 40

yeah can u see wheere this would ben true on a unit circle

Answer 41

ah im lagging cant really see my text

Answer 42

alright i think i understand it, thanks for all the help man <3

Answer 43

ye sure :)