If f (x) = x ^ 2-2 / 3x-k has zero on [- 1,1], then the value range of real number k I want a detailed process... Don't just give me an answer

The axis of symmetry of the parabola is x = 1 / 3
Let f (x) = x ^ 2-2 / 3x-k have zeros on [- 1,1]
There must be f (1 / 3) - 1 / 9
At the same time, f (- 1) > = 0 (1) or F (1) > = 0 (2)
From (1) we get K

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