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Well, firstly, the derivative of a function simply refers to slope. Usually we say that the function is not differentiable at a point.
Say you have a function such as this:
f(x)=|x|
Another way to represent that would be as a piece-wise function:
g(x) = { -x for x<0
{ x for x>= 0
The problem arises at the specific point x=0. If you look at the slope--the change in the function--from the left and right of x, you notice that it is different, negative 1 and positive 1. So, we can say that the function is not differentiable at x=0 because of that sudden change.
There are however, a few functions that are nowhere differentiable. One example is the Weirstrass function. The even more ironic thing about this function is that it is continuous everywhere! Since this function is not differentiable anywhere, many might call it a non-differentiable function.
There are absolutely other examples.
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Where is f(x) discontinuous but not differentiable Explain?
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.
How do you find critical value for a total revenue function?
If it is a differentiable function, you find the value at which its derivative is 0. But in general, you can plot it as a line graph and see where it peaks.
Is log x differentiable at 1?
yes...
Can a graph be differentiable at a specific point but not continuous at the same point?
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.
If f is a differentiable function such that f of 9 equals 2 and f' of 9 equals -2 then what is the linearization?
Use the point-slope form: y - y1 = m(x - x1). From the givens, y1 = 2, x1 = 9, and m = -2. Thus, y - 2 = -2(x - 9) = -2x + 18, or y = -2x + 20.
