Absolute Graphing (1 Viewer)

[Lukybear]

The Graphing of an absolute function has made me extremely curious.

y= |x| + 1/x

When finding the turning points, dy/dx = 0 gives x = 1 or -1

Using the second derivative, at x=1, a min t.p exist. At x=-1, a max exist. However, when graphing, and observing the graph, the max at x=-1 does not exist.

Is there something wrong with my answers which has caused this problem? Or have i missed something that I should of done? If so please enlighten.

 

[lychnobity]

The Graphing of an absolute function has made me extremely curious.

y= |x| + 1/x

When finding the turning points, dy/dx = 0 gives x = 1 or -1

Using the second derivative, at x=1, a min t.p exist. At x=-1, a max exist. However, when graphing, and observing the graph, the max at x=-1 does not exist.

Is there something wrong with my answers which has caused this problem? Or have i missed something that I should of done? If so please enlighten.

|x| + 1/x can be broken into 2 graphs:

x + 1/x (ie for x>0)
-x + 1/x (ie for x<0)

As the eqn of |x| on the LHS is -x, and is x on the RHS

So when testing for turning points <0 you should be considering the second graph, instead of taking the positive case.

 

[Lukybear]

I see. Thanks very much.

EDIT: When testing, i had used the second derivate method. When derived, the absolute x became a zero. This still allowed the point to be a maxima.

When I used the table however, it showed that indeed x=-1 was indeed not a turning point.

Any comments?

 

[lychnobity]

I see. Thanks very much.

EDIT: When testing, i had used the second derivate method. When derived, the absolute x became a zero. This still allowed the point to be a maxima.

hmm, if the first and second derivative equal zero, then it's inconclusive: you cant ascertain if it's a turning point, thus in these cases, only the table method will work.

 

[Trebla]

The Graphing of an absolute function has made me extremely curious.

y= |x| + 1/x

When finding the turning points, dy/dx = 0 gives x = 1 or -1

Using the second derivative, at x=1, a min t.p exist. At x=-1, a max exist. However, when graphing, and observing the graph, the max at x=-1 does not exist.

Is there something wrong with my answers which has caused this problem? Or have i missed something that I should of done? If so please enlighten.

For x > 0:
y = x + 1/x
dy/dx = 1 - 1/x²
Solving dy/dx = 0 yields:
x² = 1
which has solutions at x = 1 and x = -1.
HOWEVER, this function is only defined for x > 0, so we cannot consider the x = -1 case, because the function takes a different equation (y = - x + 1/x NOT y = x + 1/x) at x = -1.
Therefore you only consider the turning point at x = 1.

For x < 0:
y = - x + 1/x
dy/dx = - 1 - 1/x²
Solving dy/dx = 0 yields:
x² = - 1
which gives no real solutions, hence there should not be any turning points for x < 0

 

[Lukybear]

Very nice trebla.