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Is the derivative of the equation (-e^(-sqrt(3)x))*(sqrt(3)cos(x)+sin(x))? And when you set each factor of the derivative equal to 0 to find the horizontal tangents, is the answer -pi/3? And is it only -pi/3, or -pi/3 + pi*n? Thank you!
Date Posted: 3 years ago
Views: 324
y = e⁻√³ˣ cos(x)
Using product rule,
dy/dx = (-√3 e⁻√³ˣ) cos(x) + e⁻√³ˣ (-sin(x))
= -√3 e⁻√³ˣ cos(x) - e⁻√³ˣ sin(x)
= -e⁻√³ˣ (√3 cos(x) + sin(x))
When tangent is horizontal, dy/dx = 0
-e⁻√³ˣ (√3 cos(x) + sin(x)) = 0
-e⁻√³ˣ = 0 (rejected as it is always < 0 for all real values of x)
or √3 cos(x) + sin(x) = 0
sin(x) = -cos(x)
sin(x) / cos(x) = -√3
tan(x) = -√3
x = tan⁻¹ (-√3)
Since tan(x) is negative we consider the 2nd and 4th quadrant.
x = -π/3 , -π/3 + π , -π/3 + 2π, -π/3 + 3π,...
and -π/3 - π, -π/3 - 2π, -π/3 - 3π,...
So x = -π/3 ± πn, where n is an integer.
You are asked to find all points so there cannot be only one answer.
Using product rule,
dy/dx = (-√3 e⁻√³ˣ) cos(x) + e⁻√³ˣ (-sin(x))
= -√3 e⁻√³ˣ cos(x) - e⁻√³ˣ sin(x)
= -e⁻√³ˣ (√3 cos(x) + sin(x))
When tangent is horizontal, dy/dx = 0
-e⁻√³ˣ (√3 cos(x) + sin(x)) = 0
-e⁻√³ˣ = 0 (rejected as it is always < 0 for all real values of x)
or √3 cos(x) + sin(x) = 0
sin(x) = -cos(x)
sin(x) / cos(x) = -√3
tan(x) = -√3
x = tan⁻¹ (-√3)
Since tan(x) is negative we consider the 2nd and 4th quadrant.
x = -π/3 , -π/3 + π , -π/3 + 2π, -π/3 + 3π,...
and -π/3 - π, -π/3 - 2π, -π/3 - 3π,...
So x = -π/3 ± πn, where n is an integer.
You are asked to find all points so there cannot be only one answer.
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