Question

What is the equation of the tangent line of `f(x)=1/x-sqrt(x+1)` at `x=3`?

Answer 1

`y = -13/36x - 7/12`

Explanation:

First we begin by finding the value of `f(3)` so that we have the point of tangency locked down:

`f(3) = 1/3 - sqrt(3+1) = 1/3 - sqrt(4) = 1/3 - 2 = -5/3`

Next, in order to find the slope of the tangent at the point `(3,-5/3)` we have to have the derivative `f'(x)` so that we can evaluate it at `x=3`:

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

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

`= -1/x^2 - 1/(2(x+1)^(1/2))`

`= -1/x^2 -1/(2sqrt(x+1))`

For `x = 3`:

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

`= -1/9 - 1/(2*2) = -1/9 - 1/4 = - 13/36`

To find the equation of the tangent line, we can use the Point-Slope form of a line:

`y - y_1 = m(x-x_1)`

`y - (-5/3) = (-13/36)(x - 3)`

`y + 5/3 = -13/36x + 13/12`

`y = -13/36x + 13/12 - 5/3`

`y = -13/36x - 7/12`

Here's the graph:

graph{(1/x-sqrt(x+1)-y)((-13/36)x - 7/12 - y)=0 [-.1, 6, -3, 2.368]}

Answer 2

The equation of tangent line is `5x- 36y =75`

Explanation:

`f(x)= 1/x-sqrt(x+1) ; x=3 , f(x)= 1/3-sqrt(3+1)`

`:.f(x)= 1/3-2= -5/3 :.`The coordinates of point is

`(3,-5/3)` where tangent line is drawn. Slope of the curve

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

slope at point `x=3` is `m= -1/3^2 +1/(2sqrt(3+1)` or

`m= -1/9+1/4 or m=5/36` . The equation of tangent line

having slope of `m=5/36` paasing through point `(3,-5/3)` is

`y+5/3=5/36(x-3) or 36y +60 =5x -15` or

`5x- 36y =75` [Ans]