You need to remember that linear approximation near a point x=a is given by equation `L(x) = f(a) + f'(a)(x-a)` .

You need to find the tangent line to the curve `f(x) = 1/(sqrt(2x-1)), ` hence you need to differentiate the function with respect to x such that:

`f'(x) = (-(2x-1)')/(2sqrt(2x-1))`

`f'(x) = -2/(2sqrt(2x-1)) =gt f'(x) = -1/(sqrt(2x-1))`

You should evaluate f'(x) at x=5 such that:

`f'(5) = -1/(sqrt(10-1)) =gt f'(5) = -1/3`

You need to evaluate f(x) at x=5 such that: `f(5) = 1/3`

Hence, the linear approximation of f(x), near the point `(5,1/3)` is:

`L(x) = f(5) + f'(5)(x-5)`

`L(x) =1/3- (x-5)/3 =gt L(x) = (6-x)/3 =gt L(x) = 2 - x/3`

`L(5) = 2 - 5/3 = 1/3 = 0.33`

**Hence, evaluating the linear approximation of f(x) near the point `(5,1/3)` yields`L(5) = 0.33` .**

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