First derivative,

`f'(x) = 2/3*x^(2/3-1) -0 = 2/3x^(-1/3)`

At critical points, `f'(x) = 0`

Therefore,

`2/3x^(-1/3) = 0`

This gives, x = 0

At `x = 0, f(x) = 0-3 = -3.`

Therefore, the point (0,-3) is a critical point.

**Second derivative test**

`f''(x) = 2/3*(-1/3)*x^(-1/3-1) = -2/9x^(-4/3)`

At `x...

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First derivative,

`f'(x) = 2/3*x^(2/3-1) -0 = 2/3x^(-1/3)`

At critical points, `f'(x) = 0`

Therefore,

`2/3x^(-1/3) = 0`

This gives, x = 0

At `x = 0, f(x) = 0-3 = -3.`

Therefore, the point (0,-3) is a critical point.

**Second derivative test**

`f''(x) = 2/3*(-1/3)*x^(-1/3-1) = -2/9x^(-4/3)`

At `x = 0, f"(x) = 0 `

This means that (0,-3) is an inflection point