Question

If f(x) = 3a|4x – 4| – ax, where a is some constant, find f ′(1).
Please help me guys, I give medals @preethat @phi

Answer 1

@phi could you help me

Answer 2

I remember this question and there was something fishy about it.. if a=0 then f(x)=0 then f'(x)=0 and so f'(1)=0
but what if a is not 0
then we have to consider 4x-4>0 and 4x-4<0
if 4x-4>0 then x-1>0
and then so f(x)=3a(4)(x-1)-ax if x>1
find f' for this side

and if 4x-4<0 then x-1<0
and then so f(x)=3a(-4)(x-1)-ax if x<1
find f' for this one
see if f'(1) matches for both sides

Answer 3

the derivative of | f(x) | is f(x) f'(x)/|f(x)|

Answer 4

I say it was fishy because of the choices that came with this problem

Answer 5

something about
A) 1
B) 0
C) -1
D) can't remember

Answer 6

I don't know if those were the choices actually

Answer 7

Yeah, but could you explain me what they are actually asking, I mean I not want just the answer i was to know how to get answer because I really want to pass this exam

Answer 8

they are asking for f'(1)

Answer 9

I broken into a piecewise function

Answer 10

so that means that first I have to find the derivate function and then plug 1?

Answer 11

@phi gave you something where you don't need piecewise function

Answer 12

yes

Answer 13

f(x) = 3a|4x – 4| – ax, or
f(x)= 12a | x -1| -ax

the derivative is
12 a (x-1)/| x-1| - a

Answer 14

hopefully the a is not zero right
it should really specify that the constant is not zero

Answer 15

notice at x=1 we have an undefined quantity
and the best we can do is write a limit, which is different for 1- and 1+

Answer 16

well that already says f' doesn't exist at x=1 we don't have to do the limit thing

Answer 17

those are the only options, but I am not sure whether to pick 1 or not enough

0
not enough information
e
1

Answer 18

I would say not enough info because we don't know if a is 0 or not

Answer 19

like phi just show the f'(1) doesn't exist when a isn't 0
but when a=0 ,f(x)=0 and so f'(x)=0 and so f'(1)=0

Answer 20

you get two difference answers for two difference cases of a

Answer 21

true, I will select it and I'll see if the answer is correct, @phi said something similar, but i don't think they gave me enough information to solve the problem

Answer 22

is that a modulus or a bracket?

Answer 23

if you want me to explain the piecewise function more I can by the way @Joseluess

Answer 24

if you want

Answer 25

yes please, I am a little confused

Answer 26

modulus @14mdaz

Answer 27

\[|x|=\left[\begin{matrix}x & \text{ if } x>0 \\ -x & \text{ if } x<0 \\ 0 & \text{ if } x=0\end{matrix}\right]\]
sorry don't know how to type a pretty piecewise function
had to use the matrix code
have you ever seen this and do you understand this?

Answer 28

yes I know what you're talking about

Answer 29

|-2|=-(-2) I used row 2 since x=-2<0
|2|=2 I used row 1 since x=2>0
|0|=0 used row 3 since x=0

anyways just in case you didn't know that
now I could replace all those x's with any function and solve any resulting inequalities that occur

for example I see we have |4x-4| so
I'm going to replace all the x's with (4x-4)'s

Answer 30

\[|4x-4|=\left[\begin{matrix}4x-4 & \text{ if } 4x-4>0 \\ -(4x-4) & \text{ if } 4x-4<0 \\ 0 & \text{ if } 4x-4=0\end{matrix}\right]\]

Answer 31

now straightening this up a bit
\[|4x-4|=\left[\begin{matrix}4x-4 & \text{ if } x>1 \\ -(4x-4) & \text{ if } x<1 \\ 0 & \text{ if } x=1\end{matrix}\right] \]

Answer 32

but when x>1 what is the derivative of |4x-4| aka 4x-4 since |4x-4|=4x-4 for when x>1

Answer 33

\[(|4x-4|)'=\left[\begin{matrix}(4x-4)' & \text{ if } x>1 \\ -(4x-4)' & \text{ if } x<1 \\ ? & \text{ if } x=1\end{matrix}\right] \\ =...\]
I left that one thing as a ? because we have to see if left derivative=right derivative
(as x approaches 1 from both sides)

Answer 34

\[(|4x-4|)'=\left[\begin{matrix}(4x-4)' & \text{ if } x>1 \\ -(4x-4)' & \text{ if } x<1 \\ ? & \text{ if } x=1\end{matrix}\right] \\ (|4x-4|)'=\left[\begin{matrix}4 & \text{ if } x>1 \\ -4 & \text{ if } x<1 \\ \text{ does not exist (since*)} & \text{ if } x=1\end{matrix}\right] \\ \\ \text{ since } f'(1^+) \ \neq f'(1^-) \\ \text{ that is } 4 \neq -4 \]

Answer 35

now this already tells you that your function f(x)=3a|4x-4|-ax
if a isn't 0 that you have f'(1) doesn't exist because (|4x-4|)' evaluated at x=1 doesn't exist

but now if a=0 your function is f(x)=3(0)|4x-4|-0x=0-0=0
your function is f(x)=0
and so f'(x)=0 since derivative of a constant is 0
now f'(x)=0 does say it doesn't matter what x we have the f' value will always be 0
so f'(1)=0 if a=0

Answer 36

so summary of it all
f'(1) can be 0 or doesn't exist

Answer 37

the information that doesn't allow us to say which is all dependent on what a is

Answer 38

Okay, so that's why the answer it not enough right? okay I understand it better thanks

Answer 39

np