To determine the domain of the function \( f(x) = \cos^{-1}(2x) \), we need to consider the range of values for which the expression inside the inverse cosine function is valid. The primary condition for the inverse cosine function \( \cos^{-1}(y) \) is that its input \( y \) must be within the interval \([-1, 1]\).
Here, the input is \( 2x \), so we set up the inequality:
\(-1 \leq 2x \leq 1\)
We solve this compound inequality for \( x \) by dividing all parts of the inequality by 2:
\(-\frac{1}{2} \leq x \leq \frac{1}{2}\)
Thus, the domain of the function \( f(x) = \cos^{-1}(2x) \) is the interval:
\(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
The value of $\int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx$ is equal to
Considering the principal values of the inverse trigonometric functions, $\sin^{-1} \left( \frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1-x^2} \right)$, $-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$, is equal to
For the curve \( \sqrt{x} + \sqrt{y} = 1 \), find the value of \( \frac{dy}{dx} \) at the point \( \left(\frac{1}{9}, \frac{1}{9}\right) \).
Find the Derivative \( \frac{dy}{dx} \)
Given:\[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \]