Question

If f(x) = 3a|4x – 4| – ax, where a is some constant, find f ′(1).

a. 0
b. not enough information
c. e
d. 1

i don't understand this question because i thought the derivative doesn't exist at x=1 because there is a corner there.

Answer 1

It may or may not. Prove it.

1) Is it continuous? Does it have to be?
2) Are the left-and right- limits the same? Do they have to be?

Answer 2

first discuss lim f(x) at 1
then discuss continuity at 1

Answer 3

seems to me it is continuous at 1 since lim f =-a= f(1)

Answer 4

now you apply definition of derivative
lim f(x)-f(1)/x-1 as x goes to 1 does it exist?

Answer 5

i believe the limits are the same as x goes to 1 from the left and right side, so it does exist?

Answer 6

oh hold on i made a mistake
you have to consider x>1 and x<1
x>1 ===> f(x)=3a(4x-4)-ax
x<1 ===> f(x)=-3a(4x-4)-ax

Answer 7

now check the limits

Answer 8

an yes both limits are equal and they equal to f(1)=-a
then f must be continuous for all x

Answer 9

now it the last part i said lim f(x)-f(1)/x-1 as x goes to 1
from both sides

Answer 10

if such limit exist then f must be differential at x=1

Answer 11

let's take x>1
\[\lim_{x\to 1^{+}}\frac{3a(4x-4)-ax+a}{x-1} = \lim 11a=11a\]

Answer 12

now you try x<1 \[\lim_{x\to 1^{-}}\]

Answer 13

i can already see that the f(1) doe not exist
if you did that limit it won't be 11a

Answer 14

f'(1) i meant*

Answer 15

i'm confused as to why you subtracted a from 3a(4x-4)-ax? did you use the equation f(x+h) +f(x)?

Answer 16

oh i meant why you added not subtracted, sorry

Answer 17

yes i used the definition of derivative at 1
if f is differential at x=c, then the following limit must exist
\[\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\] this is the same as
\[\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]

Answer 18

oh okay i see what you're doing

Answer 19

what does option C says?
C.e
do you mean Does not exists

Answer 20

no i don't believe so. it just says e so i thought they were referring to Eulers number

Answer 21

oh ok
because the f'(1) does not exist
and your options don't have such thing
i don't think i made a mistake

Answer 22

yeah that was my whole thing. i thought it didn't exist so i thought i had done something wrong so i'm not sure what the question is asking

Answer 23

i gotta go
i think we did it the correct way
i don't know what the problem is

Answer 24

okay no problem. thanks for the help!