Determine the points in the interval (0, 2π) at which the graph of the function below has a horizontal tangent line.
f(x) = 10 cos x + 5 sin 2x
There should be three
Well, I don't want to be tangential here, but let's find those points! To find where the graph of the function has a horizontal tangent line, we need to find the points where the derivative of the function is equal to zero. In other words, we need to find the x-values where the slope of the function is zero.
The derivative of f(x) = 10 cos(x) + 5 sin(2x) can be found using the chain rule and some trigonometric identities. The derivative is:
f'(x) = -10 sin(x) + 10 cos(2x)
Now, let's set f'(x) equal to zero and solve for x:
-10 sin(x) + 10 cos(2x) = 0
Dividing both sides by 10, we get:
-sin(x) + cos(2x) = 0
Now, this is a bit of a tricky equation to solve analytically, so let's use a bit of cleverness (and humor) to make our lives easier. Since sin and cos are periodic functions with a period of 2π, we know that if there are any solutions in the interval (0, 2π), there will be at least one solution in the interval (0, π).
Now, let's make a substitution to make things even more interesting! Let's set a = sin(x) and b = cos(2x), and rewrite our equation:
-a + b = 0
Now, let's use a bit of humor to solve this equation. We can draw a Venn diagram!
----------------
| b |
|----------------|
| a |
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Now, the equation -a + b = 0 tells us that the two values, a and b, overlap. In other words, they intersect at the same x-values!
Now, let's look at the sine and cosine functions. We know that sin(x) and cos(x) both have maximum and minimum values of 1 and -1, respectively. Since a = sin(x), it can take on values between -1 and 1. Similarly, b = cos(2x) can also take on values between -1 and 1.
Now, let's determine when the values of a and b overlap. In order for a and b to be equal, both a and b should be equal to either 1 or -1. This happens at x = 0 and x = π.
So, the three points in the interval (0, 2π) where the graph of the function has a horizontal tangent line are x = 0, x = π/2, and x = π. Now that's some angular hilarity!
To determine the points in the interval (0, 2π) at which the graph of the function f(x) = 10 cos x + 5 sin 2x has a horizontal tangent line, we need to find the values of x where the derivative of the function is equal to zero.
1. Find the derivative of f(x) with respect to x.
f'(x) = -10 sin x + 10 cos 2x
2. Set f'(x) equal to zero and solve for x.
-10 sin x + 10 cos 2x = 0
3. Simplify the equation.
-10 sin x + 10 (1 - 2 sin^2 x) = 0
-10 sin x + 10 - 20 sin^2 x = 0
-20 sin^2 x - 10 sin x + 10 = 0
4. Factor the equation.
-10(2 sin^2 x + sin x - 1) = 0
5. Solve each factor equal to zero.
2 sin^2 x + sin x - 1 = 0
(2 sin x - 1)(sin x + 1) = 0
sin x = 1/2 or sin x = -1
6. Determine the values of x that satisfy the conditions in the interval (0, 2π).
For sin x = 1/2:
- x = π/6 or x = 5π/6
For sin x = -1:
- x = 3π/2
Thus, the points in the interval (0, 2π) at which the graph of the function has a horizontal tangent line are x = π/6, x = 5π/6, and x = 3π/2.
To find the points at which the graph of a function has a horizontal tangent line, we need to find the values of x where the derivative of the function is equal to zero or undefined.
To find the derivative of the given function f(x) = 10 cos x + 5 sin 2x, we can use the chain rule.
The derivative of cos x is -sin x, and the derivative of sin 2x is 2 cos 2x. We can then differentiate each term separately:
f'(x) = -10 sin x + 10 cos 2x · 2
Now, set f'(x) = 0 and solve for x:
-10 sin x + 10 cos 2x · 2 = 0
Dividing both sides by 20:
-0.5 sin x + cos 2x = 0
Now, we need to find the x-values that satisfy this equation within the interval (0, 2π).
One way to find these points is to graph the function y = -0.5 sin x + cos 2x and observe where it crosses the x-axis. However, this can be a bit tedious.
Alternatively, we can solve the equation analytically by using trigonometric identities.
The identity cos 2x = 1 - 2 sin^2 x can be useful here.
Substituting this into the equation:
-0.5 sin x + 1 - 2 sin^2 x = 0
Combining like terms:
-2 sin^2 x - 0.5 sin x + 1 = 0
This is now a quadratic equation in terms of sin x. To solve it, we can apply the quadratic formula:
sin x = (-b ± √(b^2 - 4ac)) / 2a
Where a = -2, b = -0.5, and c = 1. Plugging in these values:
sin x = (0.5 ± √(0.5^2 - 4(-2)(1))) / 2(-2)
Simplifying:
sin x = (0.5 ± √(0.25 + 8)) / -4
sin x = (0.5 ± √(8.25)) / -4
sin x ≈ -0.368 and 0.905
Since we are interested in the x-values within the interval (0, 2π), we take the inverse sine (sin⁻¹) of these values to find the corresponding angles between 0 and 2π:
x ≈ sin⁻¹(-0.368) ≈ 0.381 π, 1.619 π (approximately)
x ≈ sin⁻¹(0.905) ≈ 1.135 π, 1.865 π (approximately)
So, the x-values at which the graph of the function has a horizontal tangent line are:
x ≈ 0.381 π, 1.135 π, and 1.865 π (approximately)
Therefore, there are three points in the interval (0, 2π) at which the graph has a horizontal tangent line.
yeah - where f'(x) = 0
f'(x) = -10sinx + 10cos2x
now just find where that is zero