Question

Group Activity Graph the function \(f(x)=\sin ^{-1}(\sin x)\) in the viewing window \([-2 \pi, 2 \pi]\) by \([-4,4] .\) Then answer the following questions: (a) What is the domain of \(f ?\) (b) What is the range of \(f ?\) (c) At which points is \(f\) not differentiable? (d) Sketch a graph of \(y=f^{\prime}(x)\) without using NDER or computing the derivative. (e) Find \(f^{\prime}(x)\) algebraically. Can you reconcile your answer with the graph in part (d)?

Short answer

The function \(f(x)\) has domain of all real numbers and range \([-π/2, π/2]\). It is not differentiable at points \(x = (2n+1) * π/2\), where \(n\) is an integer. The graph of \(y=f'(x)\) is constant between these points with marginal discontinuities, and the derivative of \(f\) is either 1 or -1 between the discontinuity points and undefined at the points of discontinuity.

Step-by-step solution

Understanding the Function

The function given is \(f(x)=\sin^{-1}(\sin x)\), where \(\sin^{-1}\) is the inverse sine function also known as arcsin. The key characteristic of arcsin is that its output is in the range \([-π/2, π/2]\)

Graph and visuals

To get an intuitive grasp, one can graph the function in the viewing window \([-2 \pi, 2 \pi]\) by \([-4,4]\) using any graphing tool. Further, for \(-π/2 ≤ x ≤ π/2\), \(\sin(\sin^{-1}(x)) = x\), thus the graph looks like a diagonal straight line. For other values of x, it repeats the pattern.

Finding the Domain and Range

The domain of the function \(f\) is \(-∞<x<∞\), because you can plug any real number into the sine function. The range of the function is hedge bounded by \(-π/2\) and \(π/2\) due to the property of the inverse sine function. Thus, the range is \(-π/2 ≤ y ≤ π/2\)

Points of non differentiability

To find the points where the function is not differentiable, we must investigate where the function isn't continuous. As the graph was repeating every half period, the points of discontinuity (hence non differentiability) are where \(x = (2n+1) * π/2\) where \(n\) is an integer

Sketching the graph without computing the derivative

To sketch the graph of the derivative \(f'(x)\), we understand that between every two discontinuities the function is a straight line. A straight line has a constant slope which changes when we move to another segment. Therefore, the derivative must be a constant between two discontinuities (or undefined at points of discontinuity)

Finding the derivative algebraically

Algebraically, the derivative of \(f\) is found by the chain rule. \(f(x)=\sin^−1(\sin x) ⇒ f'(x) = \frac{d\sin^−1(\sin x)}{dx} = \frac{1}{√(1−\sin^2 x)} * \cos x = \cos x\). If \(x\) isn't at the points of discontinuity, then \(-π/2 ≤ x ≤ π/2\) and hence, \(\cos x \ge 0\), \(and otherwise \cos x < 0\). Thus the values shift between \(±1\). This observation is in accordance with the fact that the derivative was a constant between two discontinuity points.