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

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Answer 1 · 2017-10-18T12:44:21

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

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 · 2017-10-18T12:53:53

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

#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]