What is the tangent of f(x)=sin(2x),\at x=-pi/3 ?

The tangent of f(x)=sin(2x),\at x=-pi/3 is y=-x-(sqrt(3))/2-pi/3

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tangent of f(x)=sin(2x),\at x=-pi/3

tangent

y = - x - \frac{3}{2} - \frac{π}{3}

Solution steps

Find the tangent point:

(- \frac{π}{3}, - \frac{3}{2})

Find the slope of

f (x) = sin (2 x) : \frac{df ( x )}{d x} = cos (2 x) \cdot 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 1

Find the line with slope m=

- 1

and passing through

(- \frac{π}{3}, - \frac{3}{2}) : y = - x - \frac{3}{2} - \frac{π}{3}

y = - x - \frac{3}{2} - \frac{π}{3}

What is the tangent of f(x)=sin(2x),\at x=-pi/3 ?
The tangent of f(x)=sin(2x),\at x=-pi/3 is y=-x-(sqrt(3))/2-pi/3