What is the equation of the normal line of #f(x)=x^2/(1+4x)# at #x=-1#?

Answer 1

Equation of normal at #x=-1# is #27x+6y+29=0#

As #f(x)=x^2/(1+4x)#, at #x=-1#, we have #f(-1)=(-1)^2/(1+4(-1))=1/-3=-1/3#
And hence normal passes through #(-1,-1/3)#
Now as #f(x)=x^2/(1+4x)#,
#(df)/(dx)=((1+4x)xx2x-4xxx^2)/(1+4x)^2=((2x+8x^2)-4x^2)/(1+4x)^2#
#(df)/(dx)=(2x(1+2x))/(1+4x)^2#
and hence slope of curve i.e. tangent at #x=-1# is
#(2(-1)(1+2(-1)))/(1+4(-1))^2# or #((-2)xx(-1))/(-3)^2=2/9#
and slope of normal would be #-1/(2/9)=-9/2#
Hence, equation of normal at #x=-1# is given by #(y+1/3)=-9/2xx(x+1)#
or #6(y+1/3)=-9xx6/2xx(x+1)# or #6y+2=-27x-27# or
#27x+6y+29=0#
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Answer 2

To find the equation of the normal line of f(x) at x=-1, we need to determine the slope of the tangent line at x=-1 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can use the derivative of f(x). Taking the derivative of f(x)=x^2/(1+4x) with respect to x, we get:

f'(x) = (2x(1+4x) - x^2(4))/(1+4x)^2

Evaluating f'(-1), we substitute x=-1 into the derivative:

f'(-1) = (2(-1)(1+4(-1)) - (-1)^2(4))/(1+4(-1))^2

Simplifying the expression, we find:

f'(-1) = -6/25

The slope of the tangent line at x=-1 is -6/25.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

m_normal = -1/(f'(-1))

Substituting the value of f'(-1), we have:

m_normal = -1/(-6/25)

Simplifying, we get:

m_normal = 25/6

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation. Since the point of interest is x=-1, we can use the coordinates (-1, f(-1)).

Substituting the values into the point-slope form, we have:

y - f(-1) = (25/6)(x - (-1))

Simplifying, we get:

y - f(-1) = (25/6)(x + 1)

This is the equation of the normal line of f(x)=x^2/(1+4x) 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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