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

Answer 1

Luke Alvarez

See below

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

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

We know that derivative valuate in #x_0# 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#

Next, equation is on our line.

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

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Answer 2

Aurora Alvarado

The equation of the tangent line of f(x) = xlnx - e^(3/x) at x = 2 is y = 2ln2 - 3e^(3/2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.