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

2 Answers
Oct 18, 2017

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]}

Oct 18, 2017

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]