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

Answer 1

#100x+sqrt5y-402=0.#

Let #y=f(x)=sqrt(x/(x+1)),# so, #x=4 rArr y=f(4)=sqrt(4/5)=2/sqrt5.#
So, we reqiure the eqn. of normal to the curve : #y=f(x)=sqrt(x/(x+1)),# at the pt. #(4,2/sqrt5)#
We recall that #[dy/dx]_(x=4,y=2/sqrt5)=f'(4)# gives us the slope tgt. line to the given curve.

Now, to diff. y, we can use the Quotient Rule. Instead, have a look at these two methods :-

METHOD I :-

Taking log. of both sides of the given eqn, we get, #lny=1/2{lnx-ln(x+1)} rArr d/dx(lny)=1/2d/dx{lnx-ln(x+1)} rArr d/dy(lny)*dy/dx=1/2{1/x-1/(x+1)}=1/{2x(x+1)} rArr 1/y*dy/dx=1/{2x(x+1)} rArr dy/dx=y/{2x(x+1)}#
#:. [dy/dx]_(x=4,y=2/sqrt5)=f'(4)=(2/sqrt5)/(2*4*5)=1/(20sqrt5).#

METHOD II :-

We write the eqn. of the given curve : #y^2(x+1)=x rArr d/dxy^2(x+1)=d/dxx rArr y^2*d/dx(x+1)+(x+1)*d/dxy^2=1 rArr y^2+(x+1){d/dyy^2}dy/dx=1 rArr y^2+2y(x+1)dy/dx=1 rArr x/(x+1)+2y(x+1)dy/dx=1......, [as, y^2=x/(x+1)] rArr 2y(x+1)dy/dx=1-x/(x+1)=1/(x+1) rArr dy/dx=1/{2y(x+1)^2}#
#[dy/dx]_(x=4,y=2/sqrt5)=f'(4)=1/{2*(2/sqrt5)*25}=1/(20sqrt5),# as before!
Since normal line is perp. to tgt., its slope is #-1/(f'(4))=-20sqrt5,.# and it passes thro. pt. #(4,2/sqrt5),# its eqn. is #: y-2/sqrt5=-20sqrt5(x-4)# or, #sqrt5y-2+100x-400=0,# i.e., #100x+sqrt5y-402=0.#
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Answer 2

To find the equation of the normal line of f(x) = sqrt(x/(x+1) at x = 4, we need to determine the slope of the tangent line at x = 4 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 take the derivative of f(x) with respect to x and evaluate it at x = 4.

The derivative of f(x) = sqrt(x/(x+1) can be found using the quotient rule:

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

Simplifying this expression, we get:

f'(x) = (x+1 - x) / (2 * (x+1)^(3/2))

f'(x) = 1 / (2 * (x+1)^(3/2))

Now, we can evaluate f'(x) at x = 4:

f'(4) = 1 / (2 * (4+1)^(3/2))

f'(4) = 1 / (2 * 5^(3/2))

f'(4) = 1 / (2 * 5 * sqrt(5))

f'(4) = 1 / (10 * sqrt(5))

The slope of the tangent line at x = 4 is 1 / (10 * sqrt(5)).

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

Slope of normal line = -1 / (1 / (10 * sqrt(5)))

Slope of normal line = -10 * sqrt(5)

Now, we have the slope of the normal line. To find the equation of the normal line, we use the point-slope form of a line and substitute the values of x = 4 and the slope:

y - f(4) = (-10 * sqrt(5)) * (x - 4)

Simplifying further, we get:

y - sqrt(4/(4+1)) = (-10 * sqrt(5)) * (x - 4)

y - sqrt(4/5) = (-10 * sqrt(5)) * (x - 4)

This is the equation of the normal line of f(x) = sqrt(x/(x+1)) at x = 4.

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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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