calculus
posted by amir .
Find the intervals on which the f(x)= cos^2x is concave up and concave down.

To find if f(x) is concave up, we calculate the sign f"(x).
The interval on which f"(x)>0 is concave up, and vice versa.
Let f(x)=cos²(x),
f'(x)=2sin(x)cos(x)
f"(x)=2cos(2x)
Now plot f"(x) between 0 and 2π to get:
f"(x)≥0 for x∈[π/4,π/4]
f"(x)≤0 for x∈[π/4,3π/4]
f"(x)≥0 for x∈[3π/4,5π/4]
f"(x)≤0 for x∈[5π/4,7π/4]
...
Can you take it from here?