which points of f(x)=arcsin(sinx) are not differentiable? And what is the range of this function?
How could I do this without graphing? I am just afraid that my test, which does not allow calculators, will have some question asking me to find this information.
Also, when are functions "differentiable"? Are the cusps and corners undifferentiable? Or is it the other way around?
not differentiable if f'(x) is undefined.
f'(x) = 1/√(1-x^2)
so f' is undefined at x=±1
Now review the principal values of arcsin(x)
cusps and corners are not differentiable, since the limit of f' is not the same from both sides.
To find the points at which f(x) = arcsin(sin(x)) is not differentiable, we need to consider the properties of the inverse sine function and the sine function.
The range of the sine function is [-1, 1]. However, when we take the inverse sine of values outside this range, the function is not defined. Therefore, for f(x) = arcsin(sin(x)) to be defined, we need to ensure that sin(x) lies within the range [-1, 1].
Since sin(x) repeats its values every 2π (or 360 degrees), we can focus on finding the points of non-differentiability in a single period, such as [0, 2π].
1. Points where sin(x) = -1 or sin(x) = 1:
At x = 3π/2 and x = π/2, sin(x) = -1 and sin(x) = 1, respectively. Taking the inverse sine of these values will give us arcsin(-1) = -π/2 and arcsin(1) = π/2.
2. Points where sin(x) is undefined:
These points occur when sin(x) is outside the range [-1, 1]. In the interval [0, 2π], sin(x) is not defined for x = π/2 + nπ, where n is an integer.
Therefore, the points where f(x) = arcsin(sin(x)) is not differentiable in the interval [0, 2π] are x = 3π/2, π/2, and π/2 + nπ, where n is an integer.
Regarding the range of f(x) = arcsin(sin(x)), the inverse sine function has a range of [-π/2, π/2]. Since the sine function's range is [-1, 1], the range of f(x) = arcsin(sin(x)) will be [-π/2, π/2].
To determine this information without graphing, ensure you are familiar with the properties of inverse sine and sine functions and the intervals where they are defined. Recognizing the periodic nature of the sine function is also essential.
Regarding your second question, a function is considered differentiable at a point if its derivative exists at that point. Cusps and corners are not differentiable since the derivative does not exist at those points. In general, differentiability implies smoothness, while points of non-differentiability, like cusps or corners, indicate a lack of smoothness or sharp changes in the function.
To determine the points where the function f(x) = arcsin(sin(x)) is not differentiable, we need to examine the behavior of the inner and outer functions separately.
First, consider the inner function sin(x). This function is continuous and differentiable for all real values of x. Thus, the points of non-differentiability of f(x) = arcsin(sin(x)) are determined by the non-differentiable points of the outer function, which is arcsin(x).
The function arcsin(x) is only defined for values between -1 and 1, inclusive. Therefore, we consider the two critical values x = -1 and x = 1.
At x = -1, the function arcsin(x) has a vertical tangent line, indicating a non-differentiable point. Similarly, at x = 1, arcsin(x) also has a vertical tangent line.
Thus, the points where f(x) = arcsin(sin(x)) is not differentiable occur when sin(x) = -1 or 1, which corresponds to the critical values x = -π/2 + 2πn and x = π/2 + 2πn, where n is an integer.
Finding the range of the function f(x) = arcsin(sin(x)) without graphing can be done by understanding the properties of the arcsin function and its domain.
The range of arcsin(x) is [-π/2, π/2], which means that the resulting function f(x) will have a range of [-π/2, π/2] as well. This is because sin(x) oscillates between -1 and 1, causing arcsin(sin(x)) to produce values between -π/2 and π/2.
Regarding the concept of differentiability, a function is considered differentiable at a point if its derivative exists at that point. In other words, if the slope of the tangent line exists at that point, the function is differentiable.
Typically, functions that have cusps or corners, where the slope changes abruptly, are not differentiable at those specific points. On the other hand, continuous functions, which do not have abrupt changes in slope, are typically differentiable.
In summary, the points where f(x) = arcsin(sin(x)) is not differentiable are when sin(x) equals -1 or 1, specifically at x = -π/2 + 2πn and x = π/2 + 2πn, where n is an integer. The range of the function f(x) is [-π/2, π/2]. Cusps and corners in a function typically indicate non-differentiability, while continuous functions tend to be differentiable.