Consider the function f: [frac{1}{2}, 1] righ | vedclass

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(D) Given $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$ on $[\frac{1}{2}, 1]$.First,find the derivative: $f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} = 3\sqrt{2}(4x^2

Vedclass · Accepted Answer

Both $(I)$ and $(II)$ are correct

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(D) Given $f(x) = 4\sqrt{2}x^3 - 3\sqrt{2}x - 1$ on $[\frac{1}{2}, 1]$.First,find the derivative: $f'(x) = 12\sqrt{2}x^2 - 3\sqrt{2} = 3\sqrt{2}(4x^2 - 1)$.For $x \in [\frac{1}{2}, 1]$,$4x^2 \geq 1$,so $f'(x) \geq 0$. Thus,$f(x)$ is monotonically increasing.Evaluate endpoints: $f(\frac{1}{2}) = 4\sqrt{2}(\frac{1}{8}) - 3\sqrt{2}(\frac{1}{2}) - 1 = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} - 1 = -\sqrt{2} - 1 $f(1) = 4\sqrt{2} - 3\sqrt{2} - 1 = \sqrt{2} - 1 > 0$.Since $f(\frac{1}{2})  0$,by the Intermediate Value Theorem,there exists exactly one root in $[\frac{1}{2}, 1]$. So,$(I)$ is correct.For $(II)$,set $f(x) = 0$: $\sqrt{2}(4x^3 - 3x) = 1 \Rightarrow 4x^3 - 3x = \frac{1}{\sqrt{2}}$.Let $x = \cos 	heta$. Then $\cos(3	heta) = \cos(\frac{\pi}{4})$.$3	heta = \frac{\pi}{4} \Rightarrow 	heta = \frac{\pi}{12}$.Thus,$x = \cos \frac{\pi}{12}$ is a root. So,$(II)$ is correct.Therefore,both $(I)$ and $(II)$ are correct.

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