# Find intervals of concavity: f(x)= 2*sin x and `(0 < x < 2*pi)`

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### 2 Answers

Find the intervals of concavity for `f(x)=2sinx` :

`f'(x)=2cosx`

`f''(x)=-2sinx`

`-2sinx=0==>x=pi` on `(0,2pi)`

On `(0,pi),-2sinx<0` (Try a test value like `x=pi/2==>-2sin(pi/2)=-2`

On `(pi,2pi),-2sinx>0` (Try a test value like `x=(3pi)/2==>-2sin(3pi)/2=2`

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**The graph is concave down on `(0,pi)` and concave up on `(pi,2pi)` **

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The graph of `f(x)=2sinx` :

**Sources:**

The function `f(x) = 2*sin x`

`f'(x) = 2*cos x`

`f''(x) = -2*sin x`

For x in `(0, 2*pi)` :

`-2*sin x < 0` for x in `(0, pi)`

`-2*sin x > 0` for x in `(pi, 2*pi)`

**The function f(x) = 2*sin x is concave downwards for `(0, pi)` and concave upwards for **`(pi, 2*pi)`