y=6x-12 cos x,x in[-(pi)/(2);pi]

Given Function: We are given the function y=6x-12 cos x, and we are asked to find its domain within the interval [-(pi)/(2);pi]. The domain of a function is the set of all possible input values (x-values) for which the function is defined.Linear Function: The first part of the function, 6x, is a linear function and is defined for all real numbers. There are no restrictions on x from this part of the function.Cosine Function: The second part of the function, -12 cos x, involves the cosine function. The cosine function is defined for all real numbers, so there are no restrictions on x from this part of the function either.Domain Restrictions: Since both parts of the function are defined for all real numbers, the only restriction on the domain comes from the interval that x is allowed to be in, which is [-(pi)/(2);pi]. Therefore, the domain of the function is simply the interval [-(pi)/(2);pi].Summary: To summarize, the domain of the function y=6x-12 cos x, given the interval [-(pi)/(2);pi], is the interval itself, since both the linear and cosine parts of the function do not impose any additional restrictions on the domain.

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$y=6 x-12 \cos x, x \in\left[-\frac{\pi}{2} ; \pi\right]$

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Algebra 1

Domain and range of square root functions: equations

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y=6 x-12 \cos x, x \in\left[-\frac{\pi}{2} ; \pi\right]

Calculate derivative and critical points: Calculate the derivative of $y$ to find critical points. $y = 6x - 12 \cos x$ $\frac{dy}{dx} = 6 + 12 \sin x$ Set $\frac{dy}{dx} = 0$ to find critical points: $6 + 12 \sin x = 0$ $\sin x = -\frac{1}{2}$
Solve for x: Solve $\sin x = -\frac{1}{2}$ for $x$ in the interval $[-\frac{\pi}{2}, \pi]$ . $\newline$ $x = -\frac{\pi}{6}, \frac{7\pi}{6}$ $\newline$ But $\frac{7\pi}{6}$ is not in the interval $[-\frac{\pi}{2}, \pi]$ . $\newline$ So, the critical point is $x = -\frac{\pi}{6}$ .
Calculate second derivative: Calculate the second derivative to determine the nature of the critical point. $\newline$ $\frac{d^2y}{dx^2} = 12 \cos x$ $\newline$ Evaluate $\frac{d^2y}{dx^2}$ at $x = -\frac{\pi}{6}$ : $\newline$ $\frac{d^2y}{dx^2} = 12 \cos(-\frac{\pi}{6}) = 12 \times (\sqrt{3}/2) = 6\sqrt{3}$ $\newline$ Since 6\sqrt{3} > 0, the function has a relative minimum at $x = -\frac{\pi}{6}$ .