The function f(x) = sin^4x + cos^4x increases | vedclass

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(B) Given $f(x) = \sin^4 x + \cos^4 x$.We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2

Vedclass · Accepted Answer

$\pi/4 < x < 3\pi/8$

Vedclass · Answer

(B) Given $f(x) = \sin^4 x + \cos^4 x$.We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$.Differentiating with respect to $x$,we get $f'(x) = -\frac{1}{2} \cdot 2\sin(2x) \cdot \cos(2x) \cdot 2 = -2\sin(2x)\cos(2x) = -\sin(4x)$.The function $f(x)$ increases if $f'(x) > 0$,which means $-\sin(4x) > 0$,or $\sin(4x) This inequality holds when $\pi Dividing by $4$,we get $\pi/4 Comparing this with the given options,the interval $\pi/4 Thus,the function increases in the interval given by option $B$.

The function $f(x) = \sin^4x + \cos^4x$ increases if:

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