5) Find the points on the curve below at which the tangent is horizontal. Use n as an arbitrary integer. (Select all that apply.)
y=Sin(x) / 2+ Cos (x)
y' = cosx/2 - sinx
y'=0 where cosx = 2sinx
That is, where tanx = 1/2
So, if t = arctan(1/2) the tangent line is horizontal at
x = t + nπ
To find the points on the curve where the tangent is horizontal, we need to find the values of x where the derivative of the function is equal to zero. The derivative represents the slope of the tangent line at any given point on the curve.
Let's start by finding the derivative of the given function y = sin(x)/2 + cos(x).
Step 1: Differentiate sin(x)
The derivative of sin(x) is cos(x).
Step 2: Differentiate cos(x)
The derivative of cos(x) is -sin(x).
Step 3: Find the derivative of y
Taking the derivative of y = sin(x)/2 + cos(x), we get:
dy/dx = (cos(x))/2 - sin(x)
Now, we need to find the values of x where dy/dx = 0.
Setting the derivative equal to zero:
(dy/dx) = (cos(x))/2 - sin(x) = 0
Rearranging the equation:
cos(x)/2 = sin(x)
To solve this equation, we can use the trigonometric identity:
sin^2(x) + cos^2(x) = 1
Substituting sin^2(x) with (1 - cos^2(x)):
(1 - cos^2(x))/2 + cos^2(x) = 1/2 + cos^2(x) - (cos^2(x))/2 = 1
Rearranging:
cos^2(x)/2 = 1/2
cos^2(x) = 1
Taking the square root of both sides:
cos(x) = ±1
Now let's solve for x when cos(x) is ±1.
Case 1: cos(x) = 1
In this case, x = 2nπ, where n is an arbitrary integer.
Case 2: cos(x) = -1
In this case, x = (2n + 1)π, where n is an arbitrary integer.
Therefore, the points on the curve where the tangent is horizontal are:
x = 2nπ and x = (2n + 1)π, where n is an arbitrary integer.