Final answer:
The function f(x) = 6(cos(x))^2 - 12sin(x) has an increasing interval of (0, π/2), a local minimum of -6 at x = π/2, inflection points at x = 0 and x = π, a concave up interval of (0, π), and concave down intervals of (π/2, π) and (0, π/2).
Step-by-step explanation:
(a) Finding the interval on which f is increasing:
To determine where the function is increasing, we need to find where its derivative is positive.
Let's differentiate f(x) with respect to x:
f'(x) = -12cos(x)sin(x) - 12cos(x)
The derivative is positive when f'(x) > 0:
-12cos(x)sin(x) - 12cos(x) > 0
Factoring out -12cos(x), we get:
cos(x)(sin(x)+1) = 0
Setting each factor equal to zero, we find:
cos(x) = 0 and sin(x) = -1
The first factor cos(x) = 0 in the interval (0, π/2) gives us x = π/2.
The second factor sin(x) = -1 has no solutions in the given interval.
Therefore, the function is increasing on the interval (0, π/2).
(b) Finding the local minimum and maximum values of f:
To find the local minimum and maximum values, we need to find the critical points of the function.
The critical points occur when the derivative is equal to zero or undefined.
Setting the derivative f'(x) = -12cos(x)sin(x) - 12cos(x) equal to zero, we get:
-12cos(x)(sin(x) + 1) = 0
Since sin(x) + 1 cannot be equal to zero, the only critical point is at x = π/2.
Substituting x = π/2 into the original function, we find:
f(π/2) = 6(cos(π/2))^2 - 12sin(π/2) = 6 - 12 = -6
Therefore, the local minimum value is -6 at x = π/2.
(c) Finding the inflection points:
To find the inflection points, we need to find where the concavity of the function changes.
The second derivative of f(x) is given by:
f''(x) = 12sin(x)cos(x) - 12sin(x)
The second derivative is equal to zero when:
12sin(x)cos(x) - 12sin(x) = 0
Factoring out 12sin(x), we get:
sin(x)(cos(x) - 1) = 0
Setting each factor equal to zero, we find:
sin(x) = 0 and cos(x) = 1
The first factor sin(x) = 0 has solutions in the interval [0, π], which gives us x = 0 and x = π.
The second factor cos(x) = 1 has no solutions in the given interval.
Therefore, the inflection points are at x = 0 and x = π.
(d) Finding the interval on which f is concave up:
The function is concave up when the second derivative is positive.
Since f''(x) = 12sin(x)cos(x) - 12sin(x), we need to find where f''(x) > 0.
The second derivative is positive when sin(x) > 0:
This occurs in the interval (0, π).
(e) Finding the interval on which f is concave down:
The function is concave down when the second derivative is negative.
Since f''(x) = 12sin(x)cos(x) - 12sin(x), we need to find where f''(x) < 0.
The second derivative is negative when sin(x) < 0:
This occurs in the intervals (π/2, π) and (0, π/2).