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

Question

Isabella Ackerman · Accepted Answer

#y=-1/12(23x+xln2)+(2ln2+11)/6#

To find the equation of the tangent line of the function we can start by taking its derivative, which will give us the slope of the tangent line to the curve.

#f(x)=(x-2)/(x-3)+(lnx)/(3x)-x#

I prefer to rewrite using products rather than quotients to avoid the quotient rule, but either approach will work.

#=>f(x)=(x-2)(x-3)^-1+(lnx)(3x)^-1-x#

#=(-1)(x-2)(x-3)^-2+(x-3)^-1+(1/x)(3x)^-1+(-1)(lnx)(3x)^-2(3)-1#

#=(2-x)/(x-3)^2+1/(x-3)+1/(3x^2)-(3lnx)/(9x^2)-1#

You could choose to simplify, but since we're given a value I will save time by plugging it in now and evaluating.

#f'(2)=0-1+1/12-(3ln2)/36-1#

#=-(23+ln2)/12#

This is #m# in the point-slope form of a linear equation, #y-y_1=m(x-x_1)#.

We can find #(x_1,y_1)#, a point which lies on the tangent line. This point also lies on the path of the function, so at #x=2#, we have #f(2)=y=(ln2)/6-2#.

We can put these values into the generic equation above.

#y-((ln2)/6-2)=-(23+ln2)/12(x-2)#

#=>y=-(23+ln2)/12(x-2)+(ln2)/6-2#

#y=-((23+ln2)(x-2))/12+ln2/6-2#

#y=((-23-ln2)(x-2))/12+ln2/6-2#

#y=(-23x+46+2ln2-xln2)/12+ln2/6-2#

#y=(-23x+46+2ln2-xln2+2ln2-24)/12#

#y=(-23x+22+4ln2-xln2)/12#

#y=-1/12(23x+xln2)+(2ln2+11)/6#

Ezekiel Alexander · Answer

undefined The equation of the tangent line of f(x) at x=2 is y = -1.