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

Answer 1

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#

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#.

But keeping in mind that:

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

Now, we determine 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#.

Therefore, the formula in its standard form cannot be used, nor can the derivative be extended for continuity. However, there are still alternative approaches that we can take.

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#

The line family consists of these:

#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 the tangent lines approach the vertical line in a similar manner:

#x=-1#
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Answer 2

To find the equation of the line tangent to the function f(x) = ln(x^2+x)*(x^2+x) at x = -1, we need to find the derivative of the function and evaluate it at x = -1.

First, let's find the derivative of f(x) using the product rule and chain rule:

f'(x) = [1/(x^2+x)] * (2x+1) * (x^2+x) + ln(x^2+x) * (2x+1)

Next, we substitute x = -1 into the derivative:

f'(-1) = [1/((-1)^2+(-1))] * (2(-1)+1) * ((-1)^2+(-1)) + ln((-1)^2+(-1)) * (2(-1)+1)

Simplifying the expression:

f'(-1) = [1/2] * (-1) * 0 + ln(0) * (-1)

Since ln(0) is undefined, the derivative at x = -1 is undefined. Therefore, there is no tangent line to the function f(x) at x = -1.

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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.

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.

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.

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.

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