What is the equation of the line tangent to # f(x)=x^2 + 2/x# at # x=3#?

1 Answer
Dec 1, 2015

#y=52/9x-23/3#

Explanation:

For #f(x)=x^2 + 2/x# at # x=3#, we get #y = f(3) = 9+2/3 = 29/3#

The slope of the tangent line is given by #f'(x)#
(I will assume that you know the rules for differentiation. If you need to use the definition and need help, post a comment or another question.)

#f'(x) = 2x-2/x^2#.

So, at the point of interest, the slope of the tangent line is

#m=f'(3)=6-2/9=52/9#.

The equation of the line through #(3, 29/3)# with slope #52/9# is

#y-29/3 = 52/9(x-3)#

Which can be written in slope intercept form: #y=52/9x-23/3#