Find out the answer to "Find coordinate points where the tangent line is..."

High School
Geometry
Trigonometry
Trigonometric Functions

kejpsy

Answered question

2022-01-26

Find coordinate points where the tangent line is horizontal for

f (x) = - \sin (8 x) + 6 \cos (4 x) - 8 x

Answer & Explanation

Palandriy0

Beginner2022-01-27Added 14 answers

You have used the wrong double angle formula, it should be
$\cos 8 x = 1 - 2 \sin^{2} (4 x)$
and corresponding quadratic equation would be $2 u^{2} - 3 u - 2 = 0$ and at last you would get $\sin 4 x = - \frac{1}{2}$

pacetfv

Beginner2022-01-28Added 9 answers

Let

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

Then

g^{'} (x) = - 2 \cos 2 x - 6 \sin x - 2

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

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

= 2 (2 \sin x + 1) (\sin x - 2)

Hence the critical points of g occur when

\sin x = - \frac{1}{2}

(it is not possible for

\sin x = 2

). There are four such

x \in (- π, 2 π)

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

Consequently, the critical points of f occur when

\sin 4 x = - \frac{1}{2}

, and for

- \frac{π}{4} < x < \frac{π}{2}

, this occurs at

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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Trigonometry

Didn't find what you were looking for?