What is the equation of the tangent line of f(x) =xlnx-e^(3/x)f(x)=xlnx−e3x at x = 2x=2?

1 Answer
F. Javier B.
Jun 11, 2018

See below

Explanation:

We look for equation line y-y_0=m(x-x_0)y−y0=m(x−x0) where mm is the slope and (x_0,y_0)(x0,y0) is the passing point.

The coordinates of passing point is (2,f(2))=(2,2ln2-e^(3/2))(2,f(2))=(2,2ln2−e32)

We know that derivative valuate in x_0x0 gives the slope of tangent line, thus

f´(x)=1·lnx+x·1/x-e^(3/x)·(-3/x^2)=lnx+1+(3e^(3/x))/x^2

In x_0=2 we have f´(2)=ln2+1+3/4e^(3/2)=m

Then our line has equation

y-2ln2+e^(3/2)=(ln2+1+3/4e^(3/2))(x-2)

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