Question

If f(x) = 3a|2x – 4| – ax, where a is some constant, find f ′(2).

Answer 1

What is derivative of |2x-4| where 2x-4>0?

Answer 2

Use the definition of absolute value..
|x| is x if x>0
|x| is -x if x<0
|x| is 0 if x=0
so (|x|)'=1 when x>0
so (|x|)'=-1 when x<0
But since the right derivative as x approaches 0 doesn't equal the left derivative as x approaches 0 then (|x|)' dne exist at x=0.

Answer 3

The derivative of |2x-4| is 2 is it not? But then what is the a constant mean?

Answer 4

(|2x-4|)'=2 if 2x-4>0
(|2x-4|)'=? if 2x-4<0

Answer 5

And yes the second line there is a question to answer by you

Answer 6

Would it be -2?

Answer 7

Right so we have (|2x-4|)'=2 when x>2 and (|2x-4|)'=-2 when x<2

Answer 8

Therefore the left derivative doesn't equal the right derivative as x approaches 2

Answer 9

Therefore the derivative at x=2 doesn't exist

Answer 10

Thank you @freckles

Answer 11

What about the constant?

Answer 12

what about it?

Answer 13

Does the 3a and the -ax mean nothing to this problem to find the f'(2)?

Answer 14

if you want to find the left and right derivative

Answer 15

if x>2 then f(x)=3a(2x-4)-ax and if x<2 then f(x)=-3a(2x-4)-ax

Answer 16

\[\frac{ d }{ dx }\left| f \left( x \right) \right|=\frac{ f \left( x \right) }{ \left| f \left( x \right) \right| }f \prime \left( x \right)\]

Answer 17

So if x>2 we have f'(x)=6a-a and if x<2 then f'(x)=-6a-a

Answer 18

if a=0 then f'(2) exists and it is 0
if a=anything else then f'(2) doesn't exist

Answer 19

I put that it does not exist and it said it was wrong.

Answer 20

then it made a mistake
there is no such thing as the derivative of \(|2x-4|\) evaluated at \(2\)
there is a corner there
then again there is no such thing as a unicorn

Answer 21

\[f \prime \left( x \right)=3a \frac{ 2x-4 }{ \left| 2 x-4 \right| }\left( 2 \right)-a\]
\[f \prime \left( 2 \right)=3a \frac{ 2*2-4 }{ \left| 2*2-4 \right| }\left( 2 \right)-a\]
which shows it does not exist at x=2