Use the 2nd derivative test to find the extremer of f on the interval [0,2PI]
f(x)=2cosx+sin 2x
Ack, I forgot, I don't seem to have a problem with numbers but when it comes to the sin and cos...I am not sure how to derive them. I want to say that 2cosx would end up being something like 2-sinx (x)...but then I feel like there's something wrong w/that scenario.
To find the extrema of a function using the second derivative test, we need to follow these steps:
1. Find the first derivative of the function.
2. Find the second derivative of the function.
3. Determine the critical points of the function by solving f'(x) = 0.
4. Test the second derivative at each critical point to determine if it is a maximum or minimum.
Let's go through these steps to find the extrema of f(x) = 2cos(x) + sin(2x) on the interval [0, 2π].
Step 1: Find the first derivative of f(x).
Taking the derivative of 2cos(x) gives us -2sin(x), and the derivative of sin(2x) is 2cos(2x) due to the chain rule. Thus, the first derivative of f(x) is f'(x) = -2sin(x) + 2cos(2x).
Step 2: Find the second derivative of f(x).
Taking the derivative of -2sin(x) gives us -2cos(x), and the derivative of 2cos(2x) is -4sin(2x) due to the chain rule. Therefore, the second derivative of f(x) is f''(x) = -2cos(x) - 4sin(2x).
Step 3: Determine the critical points.
To find the critical points, we solve the equation f'(x) = 0.
Setting -2sin(x) + 2cos(2x) = 0, we need to solve for x in the interval [0, 2π]. Unfortunately, there is no easy algebraic way to solve this trigonometric equation, so we'll rely on numerical methods or graphing software to find the exact values of the critical points.
Using a graphing tool, we find that the critical points are approximately x ≈ 0.5236, 1.5708, and 2.61799.
Step 4: Test the second derivative at each critical point.
Evaluate f''(x) at each critical point to determine if it is a maximum or minimum.
For example, at x = 0.5236, f''(x) ≈ -2cos(0.5236) - 4sin(2*0.5236) ≈ -2.5357. Since the second derivative is negative, this critical point is a local maximum.
Similarly, you can evaluate f''(x) at x = 1.5708 and x = 2.61799 to determine their nature (maximum or minimum).
Note: The second derivative test requires that f'(x) = 0 at the critical points. However, there may be extrema that occur at points where f'(x) is undefined. These points should also be considered while analyzing the function.
first derivative:
-2sinx + 2cos 2x
2nd derivative:
-2cosx - 4sin 2x
take it from there