What is the slope of the line normal to the tangent line of $f(x) = sin(2x-pi/6)$ at $x= pi/3$ ?

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Answer 1 · 2017-05-18T05:29:18

No slope

Find the derivative:

$f(x)=sin(2x-pi/6)$
$f'(x)=cos(2x-pi/6)x$

Substitute $x=pi/3$ to find the slope
$f'(pi/3)=cos(2(pi/3)-pi/6)(pi/3)=cos(pi/2)(pi/3)=0$

The slope of a normal line is negative reciprocal of the derivative.
$-1/0$ is undefined then the it must be a vertical line.

What is the slope of the line normal to the tangent line of f(x) = sin(2x-pi/6) f(x) = sin(2x-pi/6) at x= pi/3 x= pi/3 ?