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

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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

Sorry for the lack of clarity in the pictures

Last edited by newstudent1234; 4 weeks ago

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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

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

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(Original post by

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

**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

Last edited by newstudent1234; 4 weeks ago

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