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

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Answer 1 · 2016-09-09T20:36:14

The Reqd. Eqn. of Tgt. is # : 11e^-3x+4y-24e^-3=0#.

We know that #f'(2)# is the slope of the tgt. to the curve

# C : y=f(x)=e^(1-x^2)/(x^2-x)# at #x=2#.

#x=2 rArr y=f(2)=1/2e^-3#.

Thus, the pt. of contact is #(2,1/2e^-3)#

We will use the Quotient Rule of Diffn. to find #f'(x)#.

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

#:. f'(x)={(x^2-x)(e^(1-x^2))(-2x)-(e^(1-x^2))(2x-1)}/(x^2-x)^2#.

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

#:. f'(2)=(e^-3(-16+8-4+1))/(4-2)^2=-11/4e^-3#.

Thus, the reqd. tgt. has slope#=-11/4e^-3# and passes through

the pt. #(2,1/2e^-3)#. Using,, the The Slope-Pt. Form, its eqn. is

#y-1/2e^-3=-11/4e^-3(x-2), or, 4y-2e^-3=-11e^-3x+22e^-3#, i.e.,

#11e^-3x+4y-24e^-3=0#.

Enjoy Maths.!