Critical numbers and using limits to explain why they don't exist at certain points

Watch
newstudent1234
Badges: 10
Rep:
?
#1
Report Thread starter 4 weeks ago
#1
So our prof used limits to explain why a critical number (?) doesn't exist at point c on the graph. Don't get the formula he used to explain it. Where did he get that from? Looks similar to the differentiating from first principles formula but that's as far as I know. Also where do the 'c's inside the function brackets disappear in the second line? Where does the 0 come from? Where do the 1s come from in the third line? Can someone explain what's going on?

Sorry for the lack of clarity in the pictures
Name:  w.JPG
Views: 5
Size:  26.1 KBName:  z.JPG
Views: 4
Size:  88.9 KB
Last edited by newstudent1234; 4 weeks ago
0
reply
mqb2766
Badges: 19
Rep:
?
#2
Report 4 weeks ago
#2
Just like your other question which was to show the function was not continuous/does not exist at a point, here youre trying to show the derivative is not continuous/does not exist at the point x=c. Just looking at the function, the gradient "jumps" at x=c, but the gradient is continuous elsewhere. So to show the derivative does not exist at that point, consider the usual limit definition of a gradient using the points x=c and x=c+h. In one case h is positive, in the other case h is negative. If the gradient was continuous at x=c, the sign of h would not matter and the two values would be the same.

In this case, when h is negative, then the gradient is positive (function increases). When h is positive, the gradient is negative (function decreases). So the gradient does not exist at x=c.

The 1s etc in the handwritten notes come from the "tent" function in the bottom left, but the text is partially obscured so I cant fully read it, but it follows the description above. I guess the function is something like
f(x) = 1 - |x|
and the critical point is x=0. The gradient is +1 when x is negative and -1 when x is positive. It jumps between these two values at x=0
Last edited by mqb2766; 4 weeks ago
1
reply
newstudent1234
Badges: 10
Rep:
?
#3
Report Thread starter 4 weeks ago
#3
(Original post by mqb2766)
Just like your other question which was to show the function was not continuous at a point, here youre trying to show the derivative is not continuous/does not exist at the point x=c. Just looking at the function, the gradient "jumps" at x=c, but the gradient is continuous elsewhere. So to show the derivative does not exist at that point, consider the usual limit definition of a gradient using the points x=c and x=c+h. In one case h is positive, in the other case h is negative. If the gradient was continuous at x=c, the sign of h would not matter and the two values would be the same.

In this case, when h is negative, then the gradient is positive (function increases). When h is positive, the gradient is negative (function decreases). So the gradient does not exist at x=c.

The 1s etc in the handwritten notes come from the "tent" function in the bottom left, but the text is partially obscured so I cant fully read it, but it follows the description above. I guess the function is something like
f(x) = 1 - |x|
and the critical point is x=0. The gradient is +1 when x is negative and -1 when x is positive. It jumps between these two values at x=0
Thank u so much for helping with that!! Got it now! Hope u have an awesome day!!
Attached files
Last edited by newstudent1234; 4 weeks ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Has your education been disrupted this academic year due to the pandemic?

Yes (124)
86.71%
No (19)
13.29%

Watched Threads

View All
Latest
My Feed