What points of the graphic should be of NON differentiability for the function | X^3 -1 | (it's in absolute value)
and a second function SQUARE ROOT OF ( | X - 1 | ) + 2 ???
I'll be happy to give 10 points for the one who helps me out with these and explain me WHY they did what they did. It's just to help me understand better all this derivative thing.
P.S: At the second function ( | x - 1 | ) is inside the square root and it's on absolute value. And "+2" is outside the square root.
and a second function SQUARE ROOT OF ( | X - 1 | ) + 2 ???
I'll be happy to give 10 points for the one who helps me out with these and explain me WHY they did what they did. It's just to help me understand better all this derivative thing.
P.S: At the second function ( | x - 1 | ) is inside the square root and it's on absolute value. And "+2" is outside the square root.
-
well your question is very well worded,so
the 1st one,
you can draw the graph and do it easily.
x cube graph lies in 1st and 3rd quadrant passing through origin.
x cube -1 will be the same graph shifted 1 unit down the y -axis towards
the negative y axis.when you take modulus of the whole thing,
so just leave the part of the graph above x axis,
then take the mirror image of the part lying below it with x
axis as the mirror.and leave the graph lying below x axis.
now the graph is in 1st and second quadrant
as you take the mirror image,there is a sharp point created
at the joining point of the new graph and old graph,
in this case its at x=1.so that is a point of non differentiability
because at that point the slop to the left that is slope of 1-x cube is -3
(just differentiate) and to right that is the slope of x cube -1 is 3.so it is a point of
non differentiability.
similarly you can draw the graph of second as well and find out.
here x=1 is the point.
hope that helps!!
the 1st one,
you can draw the graph and do it easily.
x cube graph lies in 1st and 3rd quadrant passing through origin.
x cube -1 will be the same graph shifted 1 unit down the y -axis towards
the negative y axis.when you take modulus of the whole thing,
so just leave the part of the graph above x axis,
then take the mirror image of the part lying below it with x
axis as the mirror.and leave the graph lying below x axis.
now the graph is in 1st and second quadrant
as you take the mirror image,there is a sharp point created
at the joining point of the new graph and old graph,
in this case its at x=1.so that is a point of non differentiability
because at that point the slop to the left that is slope of 1-x cube is -3
(just differentiate) and to right that is the slope of x cube -1 is 3.so it is a point of
non differentiability.
similarly you can draw the graph of second as well and find out.
here x=1 is the point.
hope that helps!!
-
i am glad you liked my answer
Report Abuse
-
This function has equation y=x^3-1 for x>1 and y=1-x^3 for x<=1.
Sketching the graph shows the cusp ax x=1 but you can see that it is differentiable
at all points for which x>=1.
The easiest way is to sketch the graph of y=(sqrt|X-1|)+2 which is the upper
half of parabola y^2=(x-1) raised by 2. You can see that it is differentiable
for x>1. For x=1 the tangent to the curve is vertical the function cannot be
differentiated at x=1, or for x<1.
Sketching the graph shows the cusp ax x=1 but you can see that it is differentiable
at all points for which x>=1.
The easiest way is to sketch the graph of y=(sqrt|X-1|)+2 which is the upper
half of parabola y^2=(x-1) raised by 2. You can see that it is differentiable
for x>1. For x=1 the tangent to the curve is vertical the function cannot be
differentiated at x=1, or for x<1.
-
Absolute function curves tend to have a cusp (sharp change of direction) at which they are non-differentiable (but not in some rare cases e.g. y = Ix^3I ). These are at the point where the absolute part of the function is zero.
For both of these that occurs at x = 1. A sketch of the curves will confirm this.
EDIT. The answer below is not correct. f(x) = 2 + sqrt(x - 1) would be the top half of a parabola raised two units. However, f(x) = 2 + sqrt( Ix - 1I ) is the top half of two parabolas joined at their apexes raised by two units. It is therefore defined and continuous over all real numbers but there is a cusp at x = 1.
For both of these that occurs at x = 1. A sketch of the curves will confirm this.
EDIT. The answer below is not correct. f(x) = 2 + sqrt(x - 1) would be the top half of a parabola raised two units. However, f(x) = 2 + sqrt( Ix - 1I ) is the top half of two parabolas joined at their apexes raised by two units. It is therefore defined and continuous over all real numbers but there is a cusp at x = 1.