Absolute value point slope question

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coooool222

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Find the slope of y = |x-4| at x = 4
Find the slope of y = |x-4| at x= -2

I am super confused about this. Isn't an absolute value function not differentiable since it doesn't have a tangent line? the limits on both side don't equal.

Or is 4 not differentiable but -2 is?
 
coooool222 said:
Find the slope of y = |x-4| at x = 4
Find the slope of y = |x-4| at x= -2
I am super confused about this. Isn't an absolute value function not differentiable since it doesn't have a tangent line? the limits on both side don't equal.
Or is 4 not differentiable but -2 is?
Click to expand...
[imath]y=|x-4|=\begin{cases} x-4&x\ge 4 \\ -x+4&x<4\end{cases}[/imath] Look at THE GRAPH
What is the slope at [imath]\bf x=-2~?[/imath]
 
The derivative at a point x=a is only concerned about a small interval around x=a. The absolute value you have is made up of two intersecting lines. All lines are differentiable everywhere except in this case where the two lines intersect. So to say that the absolute value function given is differentiable no where is totally untrue.

Please draw the function and see for yourself where it is and isn't differentiable.
 
coooool222 said:
Find the slope of y = |x-4| at x = 4
Find the slope of y = |x-4| at x= -2

I am super confused about this. Isn't an absolute value function not differentiable since it doesn't have a tangent line? the limits on both side don't equal.

Or is 4 not differentiable but -2 is?
Click to expand...
You have I believe entirely missed the point of the exercise.

[imath]y = f(x) = x^2[/imath] is differentiable everywhere, but has a slope defined as [imath]\Delta y / \Delta x[/imath]. As you correctly point out, the derivative is the slope of the line tangent to the curve or (alternatively) the limit of [imath]\Delta y / \Delta x[/imath] as [imath]\Delta x[/imath] approaches zero.

What is the line tangent to [imath]f(x) = |x - 4|[/imath] when [imath]x = 4[/imath]?

The derivative at x = 4 does not exist because the tangent does not exist (or because the relevant limit does not exist).

You have mixed up something you memorized about [imath]|x|[/imath] not being everywhere differentiable and the logical basis for the true conclusion that you memorized.

PS Obviously I agree with Stephen that saying [imath]|x - 4|[/imath] is not differentiable is wrong generally. The true conclusion is that [imath]|x - 4|[/imath] is not differentiable everywhere.
 
Last edited:
coooool222 said:
Find the slope of y = |x-4| at x = 4
Find the slope of y = |x-4| at x= -2

I am super confused about this. Isn't an absolute value function not differentiable since it doesn't have a tangent line? the limits on both side don't equal.

Or is 4 not differentiable but -2 is?
Click to expand...
The derivative of an absolute value function* is given by \(\displaystyle \frac{x}{\mid x\mid}\) so in your case \(\displaystyle \frac{d}{dx}\mid x-4\mid =\frac{x-4}{\mid x-4\mid}\) so when \(\displaystyle x = -2\) the derivative is \(\displaystyle \frac{-6}{6}\left(= -1\right)\) but when you set \(\displaystyle x = 4\) you get division by zero \(\displaystyle \left(\frac{0}{0}\right)\) hence there is no derivative at that point! (I trust you understand that division by zero is not defined?)

Graphing the function (as here) clearly shows the gradient (slope) of the function as -1 (at x=4) but, although you might think that the gradient is zero at x=4 it simply doesn't exist there due to the implicit division by zero in the derivative.

*If you want further explanation of this there's a short YouTube video you might want to watch here.
 
The Highlander said:
Graphing the function (as here) clearly shows the gradient (slope) of the function as -1 (at x=4) but, although you might think that the gradient is zero at x=4 it simply doesn't exist there due to the implicit division by zero in the derivative.
Click to expand...
"the gradient (slope) of the function as -1 (at x=4)" (as highlighted above) was, clearly, an error on my part; mea culpa!
It should, of course, had read: "the gradient (slope) of the function as -1 (at x = -2)".
 
The Highlander said:
"the gradient (slope) of the function as -1 (at x=4)" (as highlighted above) was, clearly, an error on my part; mea culpa!
It should, of course, had read: "the gradient (slope) of the function as -1 (at x = -2)".
Click to expand...
I have reported you for failing to say "mea maxima culpa."
 
Dude

Get a life. I was not even beginning to mock you. How often have you seen lookagain REPORT someone? Nor did Steven G comment on my post. Anyway, I do not need to count. I have never claimed to be a mathematician. I am just someone who remembers being a student.

But I promise you that I shall not respond to any other posts of yours. I cannot cope with someone so adolescently sensitive.
 
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JeffM said:
Dude

Get a life. I was not even beginning to mock you. How often have you seen lookagain REPORT someone? Nor did Steven G comment on my post. Anyway, I do not need to count. I have never claimed to be a mathematician. I am just someone who remembers being a student.

But I promise you that I shall not respond to any other posts of yours. I cannot cope with someone so adolescently sensitive.
Click to expand...
I thought it might be better to respond elsewhere (qv) rather than continuing in this thread with further material not relevant to it.
 
JeffM said:
I How often have you seen lookagain REPORT someone?
Click to expand...

@ JeffM -- It is wrong to address a user by "dude," unless that is his or her username.

I have not reported someone often enough. If you will do more of this reporting, especially to users who repeatedly don't show work after being told, then I would report less.

Also, once you go there by using "dude," that is part of being uncivil in the forum.

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Uneven spacing?
Sent on my cell phone
 
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