kejpsy

2022-01-26

Find coordinate points where the tangent line is horizontal for $f\left(x\right)=-\mathrm{sin}\left(8x\right)+6\mathrm{cos}\left(4x\right)-8x$

Palandriy0

Beginner2022-01-27Added 14 answers

You have used the wrong double angle formula, it should be

and corresponding quadratic equation would be

pacetfv

Beginner2022-01-28Added 9 answers

Let

$g\left(x\right)=f\left(\frac{x}{4}\right)=-\mathrm{sin}2x+6\mathrm{cos}x-2x$

Then

${g}^{\prime}\left(x\right)=-2\mathrm{cos}2x-6\mathrm{sin}x-2$

$=-2(1-2{\mathrm{sin}}^{2}x)-6\mathrm{sin}x-2$

$=2(2{\mathrm{sin}}^{2}x-3\mathrm{sin}x-2)$

$=2(2\mathrm{sin}x+1)(\mathrm{sin}x-2)$

Hence the critical points of g occur when$\mathrm{sin}x=-\frac{1}{2}$ (it is not possible for $\mathrm{sin}x=2$ ). There are four such $x\in (-\pi ,2\pi )$

$x\in \{-\frac{5\pi}{6},-\frac{\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\}$

Consequently, the critical points of f occur when$\mathrm{sin}4x=-\frac{1}{2}$ , and for $-\frac{\pi}{4}<x<\frac{\pi}{2}$ , this occurs at

$x\in \{-\frac{5\pi}{24},-\frac{\pi}{24},\frac{7\pi}{24},\frac{11\pi}{24}\}$

Then

Hence the critical points of g occur when

Consequently, the critical points of f occur when

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