determine if differentiable everywhere:
3x-abs(x)-1=y
i know you're supposed to find the derivative.... -- then my mind draws a blank? hahah please help me out
also
y=x ^ 2/5
you find the derivative and it is 2/(5x^(3/5)) now what?
thanks so much
Your function separates into two linear functions,
if x > 0 we get y = 2x -1 for a slope of 2
if x < 0 we get y = 4x -1 for a slope of 4
of course at (0,-1) we have a problem.
so we can find the derivative for all values of x, except x = 0
for y = x^(2/5)
dy/dx = (2/5)x^(-3/5)
= 2/[5x^(3/5)]
which is what you have.
What do you mean "now what?"
What do you want to do with it?
determine if differentiable everywhere:`
"everywhere" means the ℝ domain, or [-∞,∞].
If you can find a point where the function is not differentiable, then the function is not differentiable everywhere.
For a function to be "differentiable" in an interval, it needs to be continuous within the interval. This is a necessary condition.
Is the given function continuous? If not, it is not diffentiable where discontinuity occurs.
Another necessary condition for differentiability is that for every point in the interval, either
1. the derivative exists, or
2. both the left-hand and right-hand derivatives exist, and they are equal.
For
y=3x-abs(x)-1, see
http://img23.imageshack.us/img23/4836/1254714362andre1.png
y' = 3-x/abs(x) which is a step function at x=0, which means that at x=0, the derivatives from the left and right are 4 and 2 respectively. Make your conclusion whether the function is differentiable "everywhere", and if not, define the interval in which it is.
For the other function,
y=x^(2/5)
make a table of values of the function for x=-3 to +3 in steps of 1 and confirm that
1. it is continuous between -3 and +3
2. it is an even function
Also calculate y' at 0+ and 0-.
Conclude if the function is differentiable "everywhere".
See:
http://img23.imageshack.us/img23/4836/1254714362andre.png
The first link (for y = 3x-abs(x)-1 ) should read:
(Broken Link Removed)
To determine whether a function is differentiable everywhere, we need to check two things:
1. The function must be continuous everywhere within its domain.
2. The derivative of the function must exist at each point in its domain.
Let's apply these concepts to your two functions:
1. For the function 3x - |x| - 1 = y:
To find the derivative, we need to consider two cases since the absolute value function has different definitions for positive and negative values:
Case 1: x ≥ 0
In this case, |x| simplifies to just x, so the function becomes 3x - x - 1 = y, which simplifies to 2x - 1 = y.
Case 2: x < 0
In this case, |x| simplifies to -x, so the function becomes 3x - (-x) - 1 = y, which simplifies to 4x - 1 = y.
Now, let's take the derivatives of both cases:
Case 1: Differentiate 2x - 1 = y with respect to x:
dy/dx = 2
Case 2: Differentiate 4x - 1 = y with respect to x:
dy/dx = 4
Since the derivatives of both cases are constant, the function is differentiable everywhere within its domain. However, it's worth noting that the function is not differentiable at x = 0 because of the sharp corner there.
2. For the function y = x^(2/5):
The derivative you calculated, dy/dx = 2/(5x^(3/5)), is correct. Now, to determine the differentiability, we need to consider the domain of the function.
Since x^(2/5) is defined for all values of x, except x = 0, the function is continuous everywhere within its domain.
Next, we check if the derivative exists at each point in the domain. The only point where the derivative might not exist is x = 0, which is not included in the domain anyway. Hence, the function y = x^(2/5) is differentiable everywhere within its domain (excluding x = 0 in this case).
In summary, the function 3x - |x| - 1 = y is differentiable everywhere within its domain, except at x = 0. The function y = x^(2/5) is differentiable everywhere within its domain, excluding x = 0.