math homework help: Find a formula for h’(x), where h(x)= (absolute vale of x-4) + (absolute value of x-8). - Help.com

Find a formula for h’(x), where h(x)= (absolute vale of x-4) + (absolute value of x-8).

Hint: h(x) may not be differentiable at x=4 or 8.

I need h’(x) if x

This open post was written 4 months, 2 weeks ago | V/U/S: 143, 14, 6 | Edit Post | Leave a reply | Report Post


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ihicmi edited this post 4 months, 2 weeks ago. Read the previous text »

Find a formula for h’(x), where h(x)= (absolute vale of x-4) + (absolute value of x-8).

Hint: h(x) may not be differentiable at x=4 or 8.

h’(x)= blank if x

l2ndpowe changed the tags on this post: they were "" 4 months, 2 weeks ago.

Anonymous #
4 months, 2 weeks ago (4 minutes after post)

wudnt u just do…. (x-4)+(X-8)…n just get x by it self or soemthin liek that iduno

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Anonymous #
4 months, 2 weeks ago (16 minutes after post)

i dont understand.. you already have the formula h(x)= /x-4/ + /x-8/ (/ is absoluate value sign) sorry i would help but i dont get the question

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Hyperpwnage offline Verified User (6 months, 2 weeks) Long Term User Shouts: 0 #
An Unknown Location | 4 months, 2 weeks ago (17 minutes after post)

Find a formula for h’(x), where h(x)= (absolute vale of x-4) + (absolute value of x-8).

Hint: h(x) may not be differentiable at x=4 or 8.

I need h’(x) if x

Absolute value of (x-4) = (x+4)
Absolute value of (x-8) = (x+8)

H = 2x+12?

If I’m understanding the problem right… =P

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Anonymous #
4 months, 2 weeks ago (19 minutes after post)

i dont think you have the absolute value part right but i didn’t understand it either haha but the av of x-4 is not x+4

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Anonymous #
4 months, 2 weeks ago (25 minutes after post)

oh yeah absolute is the opposite sign of what the number is i think i dunno again

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Hyperpwnage offline Verified User (6 months, 2 weeks) Long Term User Shouts: 0 #
An Unknown Location | 4 months, 2 weeks ago (28 minutes after post)

Anonymous wrote:
i dont think you have the absolute value part right but i didn’t understand it either haha but the av of x-4 is not x+4

Well, absolute value of -4 is 4, so I assumed…

The absolute value of x-4 is x+4…

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Anonymous #
4 months, 2 weeks ago (28 minutes after post)

absolute value is the distance a number is from zero. absolute value of 4-5 is 1 absoluet value of 5-4 is 1 etc

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Anonymous #
4 months, 2 weeks ago (30 minutes after post)

Hyperpwnage wrote:

Anonymous wrote:
i dont think you have the absolute value part right but i didn’t understand it either haha but the av of x-4 is not x+4

Well, absolute value of -4 is 4, so I assumed…

The absolute value of x-4 is x+4…

yea but absolute value of of 4-4 is not the same of absolute value of 4+4

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Hyperpwnage offline Verified User (6 months, 2 weeks) Long Term User Shouts: 0 #
An Unknown Location | 4 months, 2 weeks ago (34 minutes after post)

>.>

Waiit. Now I’m confused…

YEH IT IS!! xD

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Hyperpwnage offline Verified User (6 months, 2 weeks) Long Term User Shouts: 0 #
An Unknown Location | 4 months, 2 weeks ago (35 minutes after post)

Yeah, the absolute value is the MAXIMUM value of the numbah. :3

I think I’m right… >.>

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l2ndpowe offline Verified User (2 years, 3 months) Long Term User Shouts: 1 #
Houston, TX, US | 4 months, 2 weeks ago (37 minutes after post)

Ihicmi, just graph it out.

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bikeminebo offline Unverified User #
An Unknown Location | 4 months, 2 weeks ago (1 hour, 11 minutes after post)

The derivative of something inside of an absolute value goes to something like this:
If y(x) = |x-a| the y’(x) = (x-a)/|x-a|.
So your answer would be : h’(x) = (x-4)/|x-4| + (x-8)/|x-8|
It is not continuous (divide by 0) at 4 or 8 so the derivative does not exist there.

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