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

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Answer 1 · 2016-08-06T17:52:48

4x+y+2+0#.

We recall that #f'(x)# is the slope of tgt. to the curve at the pt.

#(x,f(x))#

#f(x)=(x^2+1)/(x+2) rArr f(-1)=2#.

So, we require the eqn. of tgt. at the pt. #(-1,2)#

Now, #f(x)=(x^2+1)/(x+2)

#rArr (x+2)f(x)=x^2+1#

# rArr d/dx{(x+2)(f(x)}=d/dx(x^2+1)#

# rArr (x+2)*d/dx(fx)+f(x)*d/dx(x+2)=2x+1#

# rArr (x+2)*f'(x)+f(x)*1=2x#,

Letting #x=-1# in this, we get,

# (-1+2)*f'(-1)+f(-1)=-2#

# rArr f'(-1)=-2-f(-1)=-2-2=-4#

Thus, the tgt. is thro. the pt. #(-1,2)# and has slope#=-4#. Hence,

the eqn. of tgt. is # : y-2=-4(x+1)=-4x-4, i.e., 4x+y+2+0#.