Let f(x)=sin 2x + cos 2x and g(x)=x^2-1. Then | vedclass

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(A) Given $f(x) = \sin 2x + \cos 2x$ and $g(x) = x^2 - 1$. First,calculate the composite function $g(f(x))$: $g(f(x)) = (\sin 2x + \cos 2x)^2 - 1$ $g(

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$x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$

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(A) Given $f(x) = \sin 2x + \cos 2x$ and $g(x) = x^2 - 1$. First,calculate the composite function $g(f(x))$: $g(f(x)) = (\sin 2x + \cos 2x)^2 - 1$ $g(f(x)) = (\sin^2 2x + \cos^2 2x + 2 \sin 2x \cos 2x) - 1$ Since $\sin^2 	heta + \cos^2 	heta = 1$ and $2 \sin 	heta \cos 	heta = \sin 2	heta$: $g(f(x)) = (1 + \sin 4x) - 1 = \sin 4x$. $A$ function is invertible if it is one-to-one (injective) and onto (surjective) in the given domain. The function $y = \sin 	heta$ is invertible in the interval $	heta \in \left[\frac{-\pi}{2}, \frac{\pi}{2}ight]$. Here,$	heta = 4x$,so we set: $\frac{-\pi}{2} \le 4x \le \frac{\pi}{2}$ Dividing by $4$: $\frac{-\pi}{8} \le x \le \frac{\pi}{8}$. Thus,$g(f(x))$ is invertible in the domain $x \in \left[\frac{-\pi}{8}, \frac{\pi}{8}ight]$.

Let $f(x)=\sin 2x + \cos 2x$ and $g(x)=x^2-1$. Then $g(f(x))$ is invertible in the domain:

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