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

convex at

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

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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.

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

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

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

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