f(x) = 3cos(x)−cos^3(x) for 0 < x < 2π
I need help finding where it increases and decreases and where it concaves up and down.
The inflection points I have found are pi/2 and 3pi/2. Help!
To determine where the function f(x) = 3cos(x) - cos^3(x) increases and decreases, we need to find the critical points of the function. Critical points occur where the derivative of the function is equal to zero or undefined.
1. Find the derivative of f(x):
f'(x) = -3sin(x) + 3cos^2(x)sin(x)
= 3sin(x)(cos^2(x) - 1)
= -3sin(x)(1 - cos^2(x))
= -3sin(x)sin^2(x)
= -3sin(x)sin^2(x)
2. Set f'(x) = 0 and solve for x:
-3sin(x)sin^2(x) = 0
From this equation, we get two critical points:
x = 0 (since sin(0) = 0)
x = π (since sin(π) = 0)
3. Determine the intervals where the function increases or decreases:
To do this, we need to examine the sign of f'(x) in different intervals.
a) For 0 < x < π:
-3sin(x)sin^2(x) is negative as sin(x) and sin^2(x) are both positive, so f'(x) < 0.
Therefore, f(x) is decreasing on the interval 0 < x < π.
b) For π < x < 2π:
-3sin(x)sin^2(x) is positive as sin(x) is negative and sin^2(x) is positive, so f'(x) > 0.
Therefore, f(x) is increasing on the interval π < x < 2π.
Thus, f(x) decreases on the interval 0 < x < π and increases on the interval π < x < 2π.
To determine where f(x) concaves up and down, we need to find the second derivative of f(x).
4. Find the second derivative of f(x):
f''(x) = -3cos(x)sin^2(x) - 3sin^3(x)cos(x)
= -3cos(x)sin(x)(sin^2(x) + sin(x))
5. Determine the intervals where the function concaves up or down:
To do this, we need to examine the sign of f''(x) in different intervals.
a) For 0 < x < π/2:
-3cos(x)sin(x)(sin^2(x) + sin(x)) is negative as cos(x), sin(x), sin^2(x), and sin(x) are all positive.
Therefore, f''(x) < 0 and f(x) is concave down on the interval 0 < x < π/2.
b) For π/2 < x < π:
-3cos(x)sin(x)(sin^2(x) + sin(x)) is positive as cos(x) is negative and sin(x), sin^2(x), and sin(x) are positive.
Therefore, f''(x) > 0 and f(x) is concave up on the interval π/2 < x < π.
c) For π < x < 3π/2:
-3cos(x)sin(x)(sin^2(x) + sin(x)) is negative as cos(x) and sin(x) are negative and sin^2(x), and sin(x) are positive.
Therefore, f''(x) < 0 and f(x) is concave down on the interval π < x < 3π/2.
d) For 3π/2 < x < 2π:
-3cos(x)sin(x)(sin^2(x) + sin(x)) is positive as cos(x) is positive and sin(x), sin^2(x), and sin(x) are negative.
Therefore, f''(x) > 0 and f(x) is concave up on the interval 3π/2 < x < 2π.
Thus, f(x) is concave down on the intervals 0 < x < π/2 and π < x < 3π/2, and concave up on the intervals π/2 < x < π and 3π/2 < x < 2π.
In summary:
- f(x) decreases on the interval 0 < x < π.
- f(x) increases on the interval π < x < 2π.
- f(x) is concave down on the intervals 0 < x < π/2 and π < x < 3π/2.
- f(x) is concave up on the intervals π/2 < x < π and 3π/2 < x < 2π.