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

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Answer 1 · 2015-11-20T17:18:32

You can do this two ways: Newton's Linearization method, or the "normal" way.

NORMAL WAY

#f'(x) = 2x + 2/(x^2)#

#f'(1) = 2 + 2 = 4#

The tangent line must pass through #(1, ?)#, with slope #4#. What is #y# when #x = 1?#

#f(1) = 1^2 - 2/1 = -1#

So the tangent line passes through #(1,-1)#. Therefore:

#(Deltay)/(Deltax) = (y_2 - (-1))/(0 - 1) = 4#

#-4 = y_2 + 1#

#y_2 = -5#

Thus, another point on the tangent line is #(0, -5)#. The y-intercept is therefore #y = -5#, and the equation overall is #color(blue)(y = 4x - 5)#.

LINEARIZATION METHOD

#f_T(a) = f(a) + f'(a)(x-a)#

#color(blue)(f_T(1)) = f(1) + f'(1)(x-1)#

All you need is to plug in #1# to the function and its derivative, and plug it back into this formula.

#f(1) = 1^2 - 2/1 = -1#

#f'(1) = 2 + 2/1^2 = 4#

#f_T(1) = -1 + 4(x - 1)#

#= color(blue)(4x - 5)#

This one is a bit more formulaic and might require less thinking for some people. Try whichever method works for you.