Find all points on the graph of the function f(x) = 2 sin(x) + (sin(x))^2 at which the tangent line is horizontal. Consider the domain x = [0,2π).
f'(x) = 2cosx + 2sinx(cosx)
= 0 when the tangent is horizontal
2cosx(1 + sinx) = 0
cosx = 0 or sinx = -1
if cosx = 0
x = π/2 or 3π/2
f(π/2) = 2(1) + 1 = 3
f(3π/2) = 2(-1) + 1 = -1
so we have two points, (π/2 , 3) and (3π/2 , -1)
if sinx = -1
x = 3π/2 giving us the same point as above
there are two points (π/2 , 3) and (3π/2 , -1)
Thanks!!! I was getting the first point wrong!
Well, finding points on a graph where the tangent line is horizontal can be quite "uphill" task, but let's give it a shot!
To start, we need to find the derivative of the function f(x). The derivative will give us the slope of the tangent line at any given point. So, let's find the derivative:
f(x) = 2sin(x) + (sin(x))^2
Using the chain rule, we differentiate the first term as 2cos(x) and the second term as 2sin(x)cos(x):
f'(x) = 2cos(x) + 2sin(x)cos(x)
Now, to find the points on the graph where the tangent line is horizontal, we set the derivative equal to zero and solve for x:
2cos(x) + 2sin(x)cos(x) = 0
If we factor out a 2cos(x), we get:
2cos(x)(1 + sin(x)) = 0
So, we have two possibilities:
1) cos(x) = 0, which occurs when x = pi/2 and x = 3pi/2.
2) 1 + sin(x) = 0, which doesn't have any solutions in our given domain of [0,2pi).
Therefore, the only points on the graph of f(x) where the tangent line is horizontal are at x = pi/2 and x = 3pi/2.
Hope that answers your question! If not, let me know and I'll try to make it even clearer (or funnier).
To find the points on the graph of the function where the tangent line is horizontal, we need to find the x-values where the derivative of the function is equal to zero. When the derivative is zero at a specific x-value, it means that the slope of the tangent line at that point is horizontal (i.e., has a slope of zero).
First, let's find the derivative of the function f(x) = 2sin(x) + (sin(x))^2.
Taking the derivative of each term separately, we get:
f'(x) = 2cos(x) + 2sin(x)cos(x)
Now, set the derivative equal to zero and solve for x:
0 = 2cos(x) + 2sin(x)cos(x)
Divide both sides by 2cos(x):
0 = 1 + sin(x)
Subtract 1 from both sides:
sin(x) = -1
Since sin(x) equals -1 when x = 3π/2, we can conclude that the tangent line is horizontal at the point on the graph where x = 3π/2.
However, we should also consider the given domain x = [0,2π). This means we need to check if 3π/2 falls within this range. Since 3π/2 > 2π, it is outside the given domain.
Therefore, there are no points within the given domain where the tangent line is horizontal.