What is the equation of the tangent line of #f(x) =sqrt(x-15)/x-1/2# at #x=31#?

Question

1369 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-19T13:05:54

Equation of tangent is # x+ 7688 y = -2821#

# f(x) = sqrt (x-15)/x-1/2 ; x=31#

#:. f(31) = sqrt (31-15)/31-1/2=4/31-1/2=- 23/62#

Point is #(31,-23/62)# at which tangent was drawn.

# f(x) = sqrt (x-15)/x-1/2 ; #

# f^'(x) = ((x *1/(2 sqrt (x-15))- sqrt (x-15)*1))/x^2 #

Slope= #= f^'(31) = (31/8- 4)/31^2= -1/7688 #

Equation of tangent at #x=31# is #y+23/62= -1/7688(x-31)#

or # 7688 y +2852 = -x+31 #or

# x+ 7688 y = -2821#

Equation of tangent is # x+ 7688 y = -2821# [Ans]