For what values of x does the graph of f(x)=x+2sinx have a horizontal tangent?
f'(x) = 1 + 2cosx
= 0 for a horizontal tangent , (slope=0)
2cosx = -1
cosx = -1/2
x must be in quadrants II or III
and I know that cos π/3 = +1/2
x = π-π/3 or x = π + π/3
x = 2π/3 or x = 4π/3 , for 0 ≤ x ≤ 2π
Well, try not to get too twisted, but when the graph of f(x) has a horizontal tangent, it means the slope of the function is zero. In other words, the derivative of f(x) equals zero. So, to find the values of x for which the graph of f(x) has a horizontal tangent, we need to find when the derivative of f(x) equals zero. But hey, don't be scared, I'm here to make it less intimidating!
The derivative of f(x) = x + 2sin(x) is f'(x) = 1 + 2cos(x). To find when f'(x) equals zero, we solve the equation:
1 + 2cos(x) = 0.
Now, let's put on our clown noses and solve this equation step by step. Subtracting 1 from both sides, we get:
2cos(x) = -1.
Then, we divide both sides by 2, and we have:
cos(x) = -1/2.
Now, to find the values of x, we can look at the unit circle or apply some trigonometric identities, but hey, let's keep it fun and simple! When is the cosine function equal to -1/2? Well, it happens at two specific angles: π/3 and 5π/3 in the interval [0, 2π].
So, the values of x for which f(x) has a horizontal tangent are x = π/3 and x = 5π/3.
And remember, no need to get too twisted in life – sometimes, it's best to just laugh and enjoy the ride!
To find the values of x for which the graph of f(x) = x + 2sinx has a horizontal tangent, we need to find the critical points of the function where the derivative is equal to zero.
The derivative of f(x) can be found using the sum and product rule:
f'(x) = 1 + 2cosx
To find where f'(x) = 0, we need to solve the equation:
1 + 2cosx = 0
Subtracting 1 from both sides, we get:
2cosx = -1
Dividing both sides by 2, we have:
cosx = -1/2
Since the range of the cosine function is -1 to 1, we know that cosx = -1/2 has solutions. We can find these values by looking at the unit circle or using trigonometric identities.
The solutions occur at two quadrants: the second quadrant and the third quadrant.
In the second quadrant, the reference angle with cosine equal to -1/2 is π/3. Therefore, the solution in the second quadrant is:
x = π - π/3 = 2π/3
In the third quadrant, the reference angle with cosine equal to -1/2 is also π/3. However, in the third quadrant, cosine is negative. Therefore, the solution in the third quadrant is:
x = -π + π/3 = 2π/3 - π = -π/3
So, the values of x where the graph of f(x) = x + 2sinx has a horizontal tangent are x = 2π/3 and x = -π/3.
To find the values of x where the graph of f(x) = x + 2sin(x) has a horizontal tangent, we need to find the points where the derivative of f(x) is equal to zero. This is because the derivative at a point gives us the slope of the tangent line to the graph at that point, and a horizontal tangent line has a slope of zero.
Let's begin by finding the derivative of f(x). The derivative of x is simply 1, and the derivative of 2sin(x) can be found using the chain rule. The derivative of sin(x) is cos(x), so the derivative of 2sin(x) is 2cos(x).
Now, we can find the points where the derivative is equal to zero:
0 = 1 + 2cos(x)
Rearranging the equation, we get:
2cos(x) = -1
Dividing both sides by 2, we have:
cos(x) = -1/2
To find the values of x, we can take the inverse cosine (arccos) of both sides:
x = arccos(-1/2)
The cosine function has a value of -1/2 at two specific angles: π/3 and 5π/3. Therefore, we have two solutions for x:
x = π/3 and x = 5π/3
These are the values of x where the graph of f(x) = x + 2sin(x) has a horizontal tangent.