Find all relative extrema. Use the Second Derivative Test where applicable.
f(x)= cosx - x (0,2ð)
f'=-sinx-.26=0
solve for x
f"=-cosx
f' = -sinx-1
f'=0 at sinx = -1, so x = 3π/2
f" = -cosx, so f"(3π/2) = 0
Looks like an inflection point, not max or min.
So, no extrema anywhere.
To find the relative extrema of the function f(x) = cos(x) - x on the interval (0, 2π), we need to find the critical points of the function and then determine whether each critical point corresponds to a maximum, minimum, or neither.
Step 1: Find the first derivative of f(x):
f'(x) = -sin(x) - 1
Step 2: Find the second derivative of f(x):
f''(x) = -cos(x)
Step 3: Find the critical points by setting the first derivative equal to zero:
-sin(x) - 1 = 0
Adding sin(x) to both sides, we get:
sin(x) = -1
The equation sin(x) = -1 has a solution x = -π/2 in the interval (0, 2π).
Step 4: Use the Second Derivative Test to determine the nature of the critical point at x = -π/2:
Evaluate f''(x) at x = -π/2:
f''(-π/2) = -cos(-π/2) = -cos(-90°) = -cos(-π/2) = -cos(-1.57) = -0.0016 (approximately)
Since f''(-π/2) is negative, the Second Derivative Test tells us that the function has a local maximum at x = -π/2.
Therefore, the relative maximum occurs at x = -π/2.
Step 5: Evaluate the function f(x) at the critical point x = -π/2:
f(-π/2) = cos(-π/2) - (-π/2) = 0 + π/2 = π/2
So, the relative maximum of f(x) = cos(x) - x on the interval (0, 2π) occurs at x = -π/2, and the corresponding y-value is π/2.
To find the relative extrema of the function f(x) = cosx - x on the interval (0, 2π), we need to follow these steps:
1. Find the first derivative of the function f(x).
2. Set the first derivative equal to zero and solve for x to find the critical points.
3. Determine the sign changes of the first derivative around each critical point to identify if it is a local maximum or minimum.
4. Use the Second Derivative Test for the critical points to confirm if they are local maximum or minimum points.
Let's go through each step in detail:
Step 1: Find the first derivative of f(x) = cosx - x.
f'(x) = -sinx - 1
Step 2: Set the first derivative equal to zero and solve for x to find the critical points.
-sinx - 1 = 0
sinx = -1
x = -π/2
Step 3: Determine the sign changes of the first derivative around each critical point.
To determine the sign changes, we can analyze the intervals (0, -π/2) and (-π/2, 2π), and pick test values within these ranges to evaluate the sign of the first derivative:
For x = 0 (in the interval 0 to -π/2):
f'(0) = -sin(0) - 1 = -1
Since the first derivative is negative, there is a sign change from positive to negative.
For x = π (in the interval -π/2 to 2π):
f'(π) = -sin(π) - 1 = 0 - 1 = -1
Since the first derivative is negative, there is no sign change.
Step 4: Use the Second Derivative Test for the critical point x = -π/2 to confirm if it is a local maximum or minimum point.
To apply the Second Derivative Test, we need to find the value of the second derivative at x = -π/2.
f''(x) = -cosx
Evaluate f''(-π/2):
f''(-π/2) = -cos(-π/2) = 0
Since the second derivative is zero, the Second Derivative Test is inconclusive.
In conclusion, the critical point x = -π/2 is neither a local maximum nor a local minimum. Therefore, there are no relative extrema on the interval (0, 2π) for the function f(x) = cosx - x.