Is #f(x)=sinx# concave or convex at #x=(3pi)/2#?

To test if a function is concave/convex at f(a), require to find the value of f''(a)

• If f''(a) > 0 then f(x) is convex at x = a

• If f''(a) < 0 then f(x) is concave at x = a

hence f(x) = sinx

f'(x) = cosx

and f''(x) = -sinx

To determine whether (f(x) = \sin(x)) is concave or convex at (x = \frac{3\pi}{2}), we need to examine the second derivative (f''(x)) at that point.

1. Find the first derivative of (f(x) = \sin(x)) with respect to (x), (f'(x)).
2. Find the second derivative of (f(x)) with respect to (x), (f''(x)).
3. Evaluate (f''(x)) at (x = \frac{3\pi}{2}).
4. Determine the concavity/convexity based on the sign of (f''(x)) at (x = \frac{3\pi}{2}).

Given (f(x) = \sin(x)):

1. (f'(x) = \cos(x))
2. (f''(x) = -\sin(x))
3. Evaluate (f''(x)) at (x = \frac{3\pi}{2}): [f''\left(\frac{3\pi}{2}\right) = -\sin\left(\frac{3\pi}{2}\right) = -(-1) = 1]
4. Since (f''\left(\frac{3\pi}{2}\right) > 0), (f(x) = \sin(x)) is concave at (x = \frac{3\pi}{2}).