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

1 Answer
Dec 22, 2016

The tangent line cannot be found in the ordinary sense because the function is not defined for #x=-1#.

However, we can see that for #x->-1# the tangent line to the curve approaches the vertical line #x=-1#

Explanation:

In general, the equation of the tangent line to the curve #y=f(x)# in the point #(bar x, f(barx))# is given by:

#y(x) = f(barx) + f'(barx)(x-barx)#

Now, we can observe that #f(x)# is only defined for #(x^2+x) > 0#,
which means for #x in (-oo,-1) uu (0,+oo)#, so strictly speaking it is not defined in #x=-1#.

However, remembering that:

#lim_(t->0) tlnt=0#

we can remove the discontinuity and pose #f(-1) = 0#.

We now calculate the first derivative:

#f'(x) = d/(dx) [(x^2+x)ln(x^2+x)]#

#f'(x) = (x^2+x) *d/(dx) [ln(x^2+x)]+ d/(dx)[(x^2+x)]ln(x^2+x)#
#f'(x) = (x^2+x) (2x+1)/(x^2+x)+ (2x+1)ln(x^2+x)#
and finally:

#f'(x) = (2x+1)(1+ln(x^2+x))#

We can see that for #x->-1^-# (from the left, as #f(x)# is not defined for #-1 < x < 0#)

#lim_(x->-1^-) f'(x) = +oo#.

So the derivative cannot be extended for continuity and we cannot use the formula in the standard form, but we can still proceed in a different way.

Let us consider the normal to the tangent line to the curve for values of #x# approaching #x=-1# from the left in succession:

#barx_n = -1-1/n#

These are the family of lines:

#y_n(x) = f(barx_n) -1/(f'(barx_n))(x-barx_n)=-x/(f'(barx_n))+(f(barx_n)+barx_n/(f'(barx_n)))#

We can easily persuade that for #n->oo# these lines approach to the line:

#y=0#

and in the same way the tangent lines approaches the vertical line:

#x=-1#