| ln (x) |
When I put dy/dx equal to zero for the modulus of ln x I get 1/0? Even though I know there should be one at 1?
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Original post by Physics4Life
When I put dy/dx equal to zero for the modulus of ln x I get 1/0? Even though I know there should be one at 1?
Posted from TSR Mobile
Posted from TSR Mobile
the function is not differentiable at the cusp!
the tangent at that point is not unique
Original post by Physics4Life
When I put dy/dx equal to zero for the modulus of ln x I get 1/0? Even though I know there should be one at 1?
Posted from TSR Mobile
Posted from TSR Mobile
Actually, the gradient at that "cusp", at the point (1,0), isn't 0.
Original post by Physics4Life
When I put dy/dx equal to zero for the modulus of ln x I get 1/0? Even though I know there should be one at 1?
Posted from TSR Mobile
Posted from TSR Mobile
you have rearranged wrong I think..
1/x=0
swap denominator and numerator on both sides and you get
x=0
Original post by Lowerty
you have rearranged wrong I think..
1/x=0
swap denominator and numerator on both sides and you get
x=0
1/x=0
swap denominator and numerator on both sides and you get
x=0
The problem with doing that is that you are dividing by zero, and division by zero is undefined.
1/x=0 does not have a solution.
Original post by Physics4Life
When I put dy/dx equal to zero for the modulus of ln x I get 1/0? Even though I know there should be one at 1?
Posted from TSR Mobile
Posted from TSR Mobile
Are you saying that their should be a gradient at (1,0)? Interestingly, this is not the case.
To expand on what TeeEm said, you know that the derivative at a point is equal to the gradient of the tangent to the curve at that point. However if you try and draw a tangent that the point (1,0), you'll get two possible tangents. Because the tangents are different, the derivative does not exist at the point (1,0).
Original post by ThatPerson
The problem with doing that is that you are dividing by zero, and division by zero is undefined.
1/x=0 does not have a solution.
Are you saying that their should be a gradient at (1,0)? Interestingly, this is not the case.
To expand on what TeeEm said, you know that the derivative at a point is equal to the gradient of the tangent to the curve at that point. However if you try and draw a tangent that the point (1,0), you'll get two possible tangents. Because the tangents are different, the derivative does not exist at the point (1,0).
1/x=0 does not have a solution.
Are you saying that their should be a gradient at (1,0)? Interestingly, this is not the case.
To expand on what TeeEm said, you know that the derivative at a point is equal to the gradient of the tangent to the curve at that point. However if you try and draw a tangent that the point (1,0), you'll get two possible tangents. Because the tangents are different, the derivative does not exist at the point (1,0).
So that means that it has no minimum stationary point then? Even though that is the lowest point?
Original post by Physics4Life
So that means that it has no minimum stationary point then? Even though that is the lowest point?
Consider as another example the function , but where we only look at the domain . That has a minimum at 0, but it's *not* a stationary point - the function isn't even differentiable there, and it has right-derivative 1, not 0. The derivative only really makes sense "in the middle of intervals", not at endpoints.
Original post by Physics4Life
So that means that it has no minimum stationary point then? Even though that is the lowest point?
As yet another example, y = |x| has its minimum at x = 0, but the function is not differentiable at that point - if you try to draw a tangent line there you could in principle draw it at any angle to the sharp point of the graph!
Thanks for clearing this up for me guys 
you all helped a lot
you all helped a lot Original post by ThatPerson
The problem with doing that is that you are dividing by zero, and division by zero is undefined.
1/x=0 does not have a solution.
Are you saying that their should be a gradient at (1,0)? Interestingly, this is not the case.
To expand on what TeeEm said, you know that the derivative at a point is equal to the gradient of the tangent to the curve at that point. However if you try and draw a tangent that the point (1,0), you'll get two possible tangents. Because the tangents are different, the derivative does not exist at the point (1,0).
1/x=0 does not have a solution.
Are you saying that their should be a gradient at (1,0)? Interestingly, this is not the case.
To expand on what TeeEm said, you know that the derivative at a point is equal to the gradient of the tangent to the curve at that point. However if you try and draw a tangent that the point (1,0), you'll get two possible tangents. Because the tangents are different, the derivative does not exist at the point (1,0).
Ah that makes sense. Thank you
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