SOLUTION: I'm trying to find where the tangent line to a function is horizontal. The equation is y = e^(-sqrt(3)x)cos(x). I got the derivative, (-e^(-sqrt(3)x))(sqrt(3)cos(x)+sin(x)), and I

Algebra ->  Test -> SOLUTION: I'm trying to find where the tangent line to a function is horizontal. The equation is y = e^(-sqrt(3)x)cos(x). I got the derivative, (-e^(-sqrt(3)x))(sqrt(3)cos(x)+sin(x)), and I       Log On


   

Question 1179606: I'm trying to find where the tangent line to a function is horizontal. The equation is y = e^(-sqrt(3)x)cos(x). I got the derivative, (-e^(-sqrt(3)x))(sqrt(3)cos(x)+sin(x)), and I set each factor equal to zero. The first factor has no solutions, but the solution for the second one is -pi/3. I then substituted -pi/3 into the original equation to find the y coordinate for the point where the tangent is horizontal. (I got (e^(sqrt(3)pi/3)/2 for the y coordinate).
Is this correct?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

you found x%5B1%5D=-pi%2F3 and got y%5B1%5D=%28e%5E%28sqrt%283%29%28pi%2F3%29%29%29%2F2
which is a point of tangency

Secondly, find the slope of the tangent line, which is the derivative of the function, evaluated at the point: m=f′x



m=0 => if so, we have horizontal line y=a

in your case y=y%5B1%5D or

y=%28e%5E%28sqrt%283%29%28pi%2F3%29%29%29%2F2-> tangent

check if that is tangent line