# Find and equation for the line tangent to the graph of f(x)=sqrt x/6x-3, at the point (1,f(1)).

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You need to evaluate the equation of the tangent line to the graph, at the point (1,f(1)), such that:

`f(x) - f(1) = f'(1)(x - 1)`

You need to evaluate `f(1)` , hence, you need to substitute 1 for x in equation of the function, such that:

`f(1) = (sqrt 1)/(6-3) => f(1) = 1/3`

You need to evaluate f'(x) using quotient rule, such that:

`f'(x) = ((sqrt x)/(6x - 3))'`

`f'(x) = ((1/(2sqrtx))(6x - 3) - 6sqrtx)/((6x - 3)^2)`

You need to evaluate `f'(1)` , such that:

`f'(1) = ((1/2)(6 - 3) - 6sqrt1)/((6 - 3)^2)`

`f'(1) = (3/2 - 6)/9 => f'(1) = (3(1/2 - 2))/9`

Reducing duplicate factors yields:

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

You need to substitute `1/3` for `f(1)` and `-1/2` for `f'(1) ` in equation of tangent line, such that:

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

`f(x) = -x/2 + 5/6`

**Hence, evaluating the equation of the tangent line to the graph, at the point `(1,f(1))` , yields **`f(x) = -x/2 + 5/6.`