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

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) (Show Source): You can put this solution on YOUR website!

you found and got
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: =′

=> if so, we have horizontal line

in your case or

-> tangent

check if that is tangent line

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