How do you find the points that the line tangent to #f(x)=x/(x+2)# has a slope 1/2?

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Answer 1 · 2017-07-09T05:35:54

The two points are (0,0) and (-4,2).

Known fact- f'(x)/gradient = 1/2.

Thought process- what are some possible values of (x,y) when f'(x) = 1/2?

f(x) = 1 - 2/(x+2) (By long division)

f'(x) = 0 - (-1)#(x+2)^-2#(2)
= 2/#(x+2)^-2#

f'(x) = 1/2
1/2 = 2/#(x+2)^-2#
#(x+2)^2# = 4
#(x+2)# = 2 or -2
#x = 0 or -4 #

When x = 0
f(0) = 0

when x = -4
f(-4) = 2

The two points are (0,0) and (-4,2).