Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line for y=(3^1/2)x+2cos(x), 0¡Üx<2¦Ð
For a horizontal tangent dy/dx = 0
dy/dx = √3 - 2sinx = 0
2sinx = √3
sinx = √3/2
x = 60° or 120°
x = π/3 or 2π/3
when x=π/3
y = √3(π/3) + 2cos(π/3) = √3(π/3) + 2(1/2) = (π√3 + 3)/3
when x = 2π/3
y = √3(2π/3) + 2(-1/2) = 2√3 π - 3)/3
To find the point(s) at which the graph of the function has a horizontal tangent line, we need to find the values of x that make the derivative of the function equal to zero. The derivative represents the slope of the tangent line to the graph at any given point.
First, let's find the derivative of the function y = (3^1/2)x + 2cos(x). We can find the derivative by applying the rules of differentiation:
dy/dx = d/dx[(3^1/2)x] + d/dx[2cos(x)]
To differentiate (3^1/2)x with respect to x, you can treat (3^1/2) as a constant, so the derivative is (3^1/2). To differentiate 2cos(x), you can use the chain rule, which states that the derivative of cos(x) is -sin(x), and multiply it by the derivative of the inner function, which is 1.
Therefore, dy/dx = (3^1/2) - 2sin(x)
Now, to find the values of x at which the derivative is zero (dy/dx = 0), we set (3^1/2) - 2sin(x) equal to zero:
(3^1/2) - 2sin(x) = 0
2sin(x) = (3^1/2)
sin(x) = (3^1/2) / 2
Now, we need to determine the values of x between 0 and 2π that satisfy this equation. Since sin(x) is positive in the first and second quadrants, we look for the positive value of sin(x) equal to (3^1/2) / 2.
Using the inverse sine function (sin^(-1)), we find:
x = sin^(-1)((3^1/2) / 2)
x ≈ 0.786 (in radians)
So, the point(s) at which the graph of the function has a horizontal tangent line occur at x ≈ 0.786. To find the corresponding y-coordinate, we substitute this value of x back into the original function:
y = (3^1/2)(0.786) + 2cos(0.786)
Calculating this expression will give you the y-coordinate of the point(s) where the graph has a horizontal tangent line.