Question

The domain of the derivative of the function \( f(x) = \cos^{-1}(2x-5) - \sin^{-1}(x-2) \) is

• A. [2,3]
• B. \( (-\infty, 2] \cup [3, \infty) \)
• C. \( (-\infty, 2) \cup (3, \infty) \)
• D. (2,3)

Hint:

The domain of the derivative of a function can be smaller than the domain of the function itself. This typically happens with functions involving square roots or inverse trigonometric functions, where the derivative becomes undefined at the endpoints of the original domain.

Answer 1

The Correct Option is D

First, let's find the domain of the function \(f(x)\) itself.
For \( \cos^{-1}(u) \) to be defined, we must have \( -1 \le u \le 1 \).
So, \( -1 \le 2x-5 \le 1 \).
Adding 5 to all parts: \( 4 \le 2x \le 6 \).
Dividing by 2: \( 2 \le x \le 3 \). So the domain for the first term is [2, 3].
For \( \sin^{-1}(v) \) to be defined, we must have \( -1 \le v \le 1 \).
So, \( -1 \le x-2 \le 1 \).
Adding 2 to all parts: \( 1 \le x \le 3 \). So the domain for the second term is [1, 3].
The domain of the function \(f(x)\) is the intersection of the domains of its parts, which is \( [2,3] \cap [1,3] = [2,3] \).
Now, let's consider the domain of the derivative, \(f'(x)\).
The derivative of \( \cos^{-1}(u) \) is \( \frac{-u'}{\sqrt{1-u^2}} \), and the derivative of \( \sin^{-1}(v) \) is \( \frac{v'}{\sqrt{1-v^2}} \).
For the derivative to be defined, the expressions inside the square roots in the denominators must be strictly greater than zero.
For the first term's derivative: \( 1 - (2x-5)^2>0 \).
\( (2x-5)^2<1 \implies -1<2x-5<1 \).
\( 4<2x<6 \implies 2<x<3 \).
For the second term's derivative: \( 1 - (x-2)^2>0 \).
\( (x-2)^2<1 \implies -1<x-2<1 \).
\( 1<x<3 \).
The domain of the derivative \(f'(x)\) is the intersection of these two open intervals.
Intersection of \( (2,3) \) and \( (1,3) \) is \( (2,3) \).