The absolute value of x, |x|, is defined as |x| = x, x>=0; -x, x<0. If you derive this, then you will find that the derivative is 1 when x>=0, and -1 when x<0. But this means that the derivative as x approaches 0 from the left does not equal the derivative as x approaches 0 from the right, as -1=/=1. So the limit as x approaches 0 does not exist, and therefore the gradient does not exist at that point, and so |x| cannot be differentiated at x = 0.
No. The absolute value is non-negative but, to be pedantic, that does not mean positive. The absolute value of 0 is 0 which is NOT positive.
The absolute value of something is its difference from 0. Absolute value of -607=607
It is the magnitude, or absolute value, of the real number.
Th absolute value of a number is its distance, regardless of direction, form 0.
Numerical value = absolute value For any real number a, the absolute value of a, denoted by |a| is itself if a ≥ 0, and -a if a < 0. Thus |a| is positive expect when a = 0.
Wherever a function is differentiable, it must also be continuous. The opposite is not true, however. For example, the absolute value function, f(x) =|x|, is not differentiable at x=0 even though it is continuous everywhere.
The distance of a point from 0 is called its magnitude or absolute value. It is the measure of how far the point is from the origin on a number line or coordinate system.
Not according to the usual definitions of "differentiable" and "continuous".Suppose that the function f is differentiable at the point x = a.Then f(a) is defined andlimit (h -> 0) [f(a+h) - f(a)]/h exists (has a finite value).If this limit exists, then it follows thatlimit (h -> 0) [f(a+h) - f(a)] exists and equals 0.Hence limit (h -> 0) f(a+h) exists and equals f(a).Therefore f is continuous at x = a.
Let f(x)=abs(x) , absolute value of x defined on the interval [5,5] f(x)= |x| , -5 ≤ x ≤ 5 Then, f(x) is continuous on [-5,5], but not differentiable at x=0 (that is not differentiable on (-5,5)). Therefore, the Mean Value Theorem does not hold.
Definition: A function f is differentiable at a if f'(a) exists. it is differentiable on an open interval (a, b) [or (a, ∞) or (-∞, a) or (-∞, ∞)]if it is differentiable at every number in the interval.Example: Where is the function f(x) = |x| differentiable?Answer:1. f is differentiable for any x > 0 and x < 0.2. f is not differentiable at x = 0.That's mean that the curve y = |x| has not a tangent at (0, 0).Thus, both continiuty and differentiability are desirable properties for a function to have. These properties are related.Theorem: If f is differentiable at a, then f is continuous at a.The converse theorem is false, that is, there are functions that are continuous but not differentiable. (As we saw at the example above. f(x) = |x| is contionuous at 0, but is not differentiable at 0).The three ways for f not to be differentiable at aare:a) if the graph of a function f has a "corner" or a "kink" in it,b) a discontinuity,c) a vertical tangent
the answer is 0. absolute value changes negitives to positives
0
No. The absolute value is non-negative but, to be pedantic, that does not mean positive. The absolute value of 0 is 0 which is NOT positive.
The absolute value of something is its difference from 0. Absolute value of -607=607
The absolute value of a number is the distance from that number to 0. Therefore, the absolute value is ALWAYS positive. the absolute value of -4.2 is 4.2 To find the absolute value, just determine how far it is from 0.
The absolute value of a number can be considered as the distance between 0 and that number on the real number line. example. or l2+6l=8 the point is to solve in between the absolute value lines
The absoulte value of Zero is 0