# What is equation of tangent to curve f(x)=(2x-1)^1/3-(2x+1)^1/3 in x=0?

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### 1 Answer

You need to write the equation of tangent line to the graph of a function `f(x)` , at a given point `(x_0,y_0)` , such that:

`y - y_0 = f'(x_0)(x - x_0)`

Since the problem gives the coordinate `x_0 = 0` , you may evaluate `y_0` coordinate replacing `0` for `x` in equation of the function, such that:

`f(0) = root(3)(2*0 - 1) - root(3)(2*0 + 1)`

`f(0) = -1 - 1 = -2`

You need to evaluate `f'(x)` , hence, differentiating the function with respect to `x` , yields:

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

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

`f'(x) = (2/3)(1/(root(3)((2x-1)^2)) - 1/(root(3)((2x+1)^2)))`

You need to evaluate `f'(x)` at `x = 0` , such that:

`f'(0) = (2/3)(1/(root(3)((0-1)^2)) - 1/(root(3)((0+1)^2)))`

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

`f'(0) = (2/3)*0`

`f'(0) = 0`

Hence, replacing `0` for `x_0, -2` for `y_0` and 0 for `f'(x_0)` , yields:

`y - (-2) = 0*(x - 0) => y + 2 = 0 => y = -2`

**Hence, evaluating the equation of the tangent line to the graph of the function `f(x) = root(3)(2x - 1) - root(3)(2x + 1)` , at `x = 0` , yields `y = -2` .**