Consider the function below. (Round the answers to two decimal places.
f(x) = 2x tan(x)
-p/2 < x < p/2
(a) Find the interval where the function is increasing.
Find the interval where the function is decreasing.
(b) Find the local minimum value.
(c) Find the interval where the function is concave up.
To find the intervals where the function is increasing or decreasing, we need to examine the sign of the derivative of the function. The derivative of f(x) with respect to x can be found by applying the product rule:
f'(x) = 2 * tan(x) + 2x * sec^2(x)
First, let's find the interval where the function is increasing:
Step 1: Find the critical points by setting the derivative equal to zero and solving for x. In this case, there are no critical points since the derivative does not equal zero within the given interval.
Step 2: Determine the behavior of the function on the intervals between the critical points and at the endpoints of the given interval.
Since there are no critical points, let's examine the behavior at the endpoints:
For x = -π/2 (lower endpoint):
Evaluate f'(-π/2) = 2 * tan(-π/2) + 2(-π/2) * sec^2(-π/2) = 2 * (-∞) + 2 * (-π/2) * (∞) = (-∞) + π * (∞) = (-∞)
For x = π/2 (upper endpoint):
Evaluate f'(π/2) = 2 * tan(π/2) + 2(π/2) * sec^2(π/2) = 2 * (∞) + 2 * (π/2) * (∞) = (∞) + π * (∞) = (∞)
Since the derivative is negative at the lower endpoint and positive at the upper endpoint, we conclude that the function is increasing on the interval (-π/2, π/2).
Next, let's find the interval where the function is decreasing:
Since we found earlier that the function is increasing on the interval (-π/2, π/2), the function is decreasing on the complement of that interval, which is (-∞, -π/2] U [π/2, ∞).
Moving on to the next part:
To find the local minimum value, we need to locate the critical points of the function within the given interval (-π/2, π/2). Since the derivative does not equal zero within this interval, there are no critical points and therefore, no local minimum value exists.
For the last part:
To find the interval where the function is concave up, we need to examine the sign of the second derivative of the function. The second derivative of f(x) with respect to x can be found by differentiating the derivative obtained earlier:
f''(x) = 2 * sec^2(x) + 2 * sec^2(x) + 4x * sec(x) * tan(x)
Next, let's determine the sign of the second derivative within the given interval (-π/2, π/2):
For any value of x within the interval (-π/2, π/2), the term 2 * sec^2(x) is always positive.
For the term 4x * sec(x) * tan(x), let's examine its sign at the endpoints:
For x = -π/2 (lower endpoint):
Evaluate 4(-π/2) * sec(-π/2) * tan(-π/2) = 4(π/2) * (∞) * (-∞) = (-∞)
For x = π/2 (upper endpoint):
Evaluate 4(π/2) * sec(π/2) * tan(π/2) = 4(π/2) * (∞) * (∞) = (∞)
Since the second derivative is negative at the lower endpoint and positive at the upper endpoint, we conclude that the function is concave up on the interval (-π/2, π/2).
In summary:
(a) The function is increasing on the interval (-π/2, π/2).
(b) There is no local minimum value.
(c) The function is concave up on the interval (-π/2, π/2).