How do you find the equation of a line tangent to the function #y=x+4/sqrtx# at (1,5)?

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Answer 1 · 2016-12-14T05:15:16

#y=-x+6#

#y=x+4/sqrtx#

#= x+4x^(-1/2)#

#y' = d/dx(x+4x^(-1/2))#

#=1-2x^(-3/2) = 1-2/sqrt(x^3)#

The equation of a straight line through #(x_1, y_1)# is:

#(y-y_1) = m(x-x_1)# Where m is the slope the line

The slope of #y# at #(1, 5)# is #y'(1) = 1-2/1 = -1#

Hence the slope of the tangent to #y# at #(1, 5)# is:

#(y-5) = -1(x-1)#

#y=-x+6#
graph{(y-x-4/sqrtx)(y+x-6)=0 [-7.04, 12.96, -1.04, 8.96]}