62​–​65. {Use of Tech} Graphing f and f'
c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.
f(x)=(x²−1)sin^−1 x on [−1,1]

Verified step by step guidance

1. Step 1: Understand the problem. We need to verify that the zeros of the derivative of the function f(x) = (x^2 - 1)sin^{-1}(x) correspond to points where the original function f(x) has a horizontal tangent line.
2. Step 2: Find the derivative f'(x). Use the product rule for differentiation, since f(x) is a product of two functions: u(x) = x^2 - 1 and v(x) = sin^{-1}(x). The product rule states that (uv)' = u'v + uv'.
3. Step 3: Differentiate u(x) and v(x) separately. For u(x) = x^2 - 1, the derivative u'(x) = 2x. For v(x) = sin^{-1}(x), the derivative v'(x) = 1/√(1-x^2).
4. Step 4: Apply the product rule. Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula to find f'(x) = (x^2 - 1)(1/√(1-x^2)) + (2x)(sin^{-1}(x)).
5. Step 5: Set f'(x) = 0 to find the zeros of the derivative. Solve the equation (x^2 - 1)(1/√(1-x^2)) + (2x)(sin^{-1}(x)) = 0 to find the x-values where f'(x) is zero. These x-values correspond to points where f(x) has a horizontal tangent line.