- #1

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So my function is y=|x+4|

1) y^2=x+4

2)2y(dy/dx)=1

3)dy/dx = 1/2y

4)dy/dx = 1/2|x+4|

I set that 0 and get

0=1/(2(|x+4|))

Am i write in thinking this cannot be solved? or missing something?

Thanks

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- Thread starter ojsimon
- Start date

- #1

- 56

- 0

So my function is y=|x+4|

1) y^2=x+4

2)2y(dy/dx)=1

3)dy/dx = 1/2y

4)dy/dx = 1/2|x+4|

I set that 0 and get

0=1/(2(|x+4|))

Am i write in thinking this cannot be solved? or missing something?

Thanks

- #2

Mentallic

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- #3

Cyosis

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It goes wrong from the start, [itex]y^2 \neq x+4[/itex].

- #4

Mentallic

Homework Helper

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When that mistake is fixed, you'll still find the the derivative cannot equal zero anywhere and it's still undefined at x=-4 with the form 0/0

- #5

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Oh yeah, thanks,

- #6

Mark44

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y = x + 4, x >= -4

y = -(x + 4), x < -4

Then y' = 1 for x > -4 and y' = -1 for x < - 4. y' does not exist at x = -4.

Any extreme points of a function occur at places where y' = 0, or y' is undefined, or at finite endpoints of the domain in cases where a function is defined only on an interval [a, b].

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