Question

g(x)=-x+4
f(x)=-2(abs(x-2))+4
Find g(f(x)) g(x)=-x+4
f(x)=-2(abs(x-2))+4
Find g(f(x)) @Mathematics

Answer 1

g(f(x)) = -2|x-2|+4+4 = -2|x-2|+8

Answer 2

Oh cool that is the answer I got. Thanks hero

Answer 3

Well, actually

Answer 4

What?

Answer 5

Oh ya i see that

Answer 6

There's a negative in front of x

Answer 7

-(-2|x-2|+4)+4

Write it like that instead

Answer 8

But lets say I have to find the derivative it would be easier to open the brackets

Answer 9

Well, the absolute value kind of complicates things

Answer 10

Lets say the derivative of v'(1)

Answer 11

Well, if that's the case, then it is probably zero

Answer 12

alright I will use the site you gave me yesterday but I don't know how I will show absolute value

Answer 13

once you replace x with a number, and evaluate it, then the derivative of any number is just zero

Answer 14

Oh so maybe I should defferentiate it first and then replace the x value with 1

Answer 15

hold the phone. it is true that the derivative of a constant is zero. but that means the derivative of a constant function is 0

Answer 16

so for example the derivative of 4 is 0, but the derivative of
\[f(x)=x^2\] at x = 4 is
\[f'(4)=2\times 4=8\] not 0!

Answer 17

ya true

Answer 18

so if you want
\[v'(1) \] the first thing you need is
\[v'(x)\]

Answer 19

and for that the first thing you need is
\[v(x)\] without absolute values signs

Answer 20

so you have
\[v(x)=4|x-2|+8\] i believe. and what this is depends on whether x is greater than or less than 2

Answer 21

Where'd you get the 4 from?

Answer 22

Oh ya? that is cool!!!!

Answer 23

if x is greater than 2 you get
\[v(x)=2(x-2)+8=2x\] whereas if x is less than 2 you get
\[v(x)=2(2-x)+8=12-2x\]

Answer 24

damn two typos

Answer 25

@hero the 4 is a straight up mistake. it is 2, not 4

Answer 26

and also
\[2(x-2)+8=2x+4\]

Answer 27

Yeah, but like you said, it's not the only one

Answer 28

in any case since 1 is less than 2, use the second formula, and see that the derivative is -2

Answer 29

oh great! Thanks for your help

Answer 30

There's still another mistake lurking

Answer 31

If I'm not mistaken, distribution comes before addition

Answer 32

yes it does!!!

Answer 33

v(x) = -(-2|x-2|+4)+4

Answer 34

I'm wondering if that simplifies to v(x) = 2|x-2|

Answer 35

I don't know

Answer 36

I guess satelliate gave up

Answer 37

I guess so ;D

Answer 38

If we continue with his logic, it splits into two:

v(x) = 2(x-2)
v(x) = 2(2-x)

Answer 39

v(x) = 2x-4
v(x) = 4 - 2x

v'(x) = 2
v'(x) = -2

Answer 40

I don't know which one we use though

Answer 41

I will play aound with it and see what I get

Answer 42

It is -2 I just calculated it on my graphing calc

Answer 43

Good

Answer 44

But i am confused as to how you got the equation of the derivative

Answer 45

Did you use the chain rule

Answer 46

Yes, why?

Answer 47

You don't know chain rule yet?

Answer 48

Ya i do

Answer 49

Hey look the derivative we got was totally wrong! That is interesting

Answer 50

Oh well. At least you found the right one

Answer 51

Actually, it's still 2

Answer 52

Take a closer look

Answer 53

sqrt and squared cancel in denominator

(x-2) cancels top and bottom

Answer 54

:D

Answer 55

Oh I will check that out

Answer 56

i am sure you are long gone, but just for the record that is not 2. it is 2 if x is greater than 2, -2 if x is less than 2. just thought i would mention it.

Answer 57

what?

Answer 58

Oh I will check that out

Answer 59

hero wrote out the correct answer above
he wrote
v(x) = 2x-4
v(x) = 4 - 2x

v'(x) = 2
v'(x) = -2

Answer 60

the only thing missing is for which values of x you use the different formulas. it is

v(x) = 2x-4 if x > 2
v(x) = 4 - 2x if x < 2

v'(x) = 2 if x > 2
v'(x) = -2 if x < 2

Answer 61

and therefore
\[v'(1)=-2\] whereas
\[v'(5)=2\]

Answer 62

I graphed the derivative and it only shows a straight line starting at x=2 until positive infinity at y=2

Answer 63

Ya you are correct

Answer 64

the graph of the derivative is two horizontal lines

Answer 65

oh great thanks