determine the open intervals on which f(x) is increasing, decreasing, concave up and concave down.
f(x)= tanx, -pi/2< x< pi/2
nevermind i figured it out
Good. I was going to suggest you just plot the function.
It increases where the derivative is positive, and is concave upward where the second derivative is positive. tan x is always increasing in that interval
To determine the intervals on which a given function is increasing, decreasing, concave up, or concave down, we need to analyze the first and second derivatives of the function.
Let's start by finding the first derivative of f(x):
f'(x) = d/dx(tan(x))
Using the chain rule, we can rewrite this as:
f'(x) = sec^2(x)
To determine where the function is increasing or decreasing, we need to find the critical points, which occur where the first derivative is either zero or undefined.
To find where f'(x) is zero, we need to solve the equation:
sec^2(x) = 0
Since sec^2(x) is always positive and never equal to zero, there are no critical points where f'(x) is zero.
Next, let's find where f'(x) is undefined. This happens when sec^2(x) is undefined, which occurs when cos(x) = 0. In the given interval (-π/2 < x < π/2), cos(x) is never zero. Therefore, there are no points where f'(x) is undefined.
Now, we can determine the intervals where f(x) is increasing or decreasing:
Since f'(x) = sec^2(x) is always positive in the given interval (-π/2 < x < π/2), this means that f(x) = tan(x) is always increasing on this interval.
To determine the concave up and concave down intervals, we need to find the second derivative of f(x). Let's calculate the second derivative:
f''(x) = d/dx(sec^2(x))
Using the chain rule and the derivative of sec^2(x) = tan^2(x) + 1, we get:
f''(x) = 2sec^2(x)tan(x)
To find where f''(x) is zero or undefined, we need to solve the equation:
2sec^2(x)tan(x) = 0
Since neither sec^2(x) nor tan(x) can be zero, there are no critical points where f''(x) is zero.
To determine where f''(x) is undefined, we need to find where sec^2(x) is undefined. This occurs when cos(x) = 0. In the given interval (-π/2 < x < π/2), cos(x) is never zero. So, there are no points where f''(x) is undefined.
Now, let's determine the intervals where f(x) is concave up or concave down:
Since f''(x) = 2sec^2(x)tan(x) is always positive in the given interval (-π/2 < x < π/2), this means that f(x) = tan(x) is always concave up on this interval.
In summary:
- f(x) = tan(x) is always increasing on the interval (-π/2 < x < π/2).
- f(x) = tan(x) is always concave up on the interval (-π/2 < x < π/2).