What is the equation of the normal line of #f(x)=((5-x)(4-x^2))/(x^3-1)# at #x=3#?

Answer 1

#y=-7.6bar(81)x+22.66#

#f(x)=((5−x)(4−x^2))/(x^3−1);quadquadquada=3#
normal line: #quady=f(a)-1/(f'(a))(x-a)#
#color(blue)(f(3)=)((5−3)(4−(3)^2))/(3^3−1)=(2*(-5))/(26)=color(blue)(-5/13)~~-0.384#
Let: #g(x)=(5−x)(4−x^2)# Then: #color(red)(g'(x)=)-1(4-x^2)+(5-x)*(-2x)=color(red)(-(4-x^2)-2x(5-x))#
#f'(x)=([-(4-x^2)-2x(5-x)]xx(x^3−1)-(5−x)(4−x^2)xx3x^2)/(x^3−1)^2#

Don't bother to simplify. Just substitute x for 3

#f'(3)=([-(4-9)-6(2)]xx(26)-(2)(-5)xx27)/(26)^2#
#f'(3)=([+5-12]xx(26)+10xx27)/(26)^2=(-7xx26+270)/26^2#
#color(orange)(f'(3)=)(-182+270)/676=88/676=44/338=color(orange)(22/169)~~0.13#
#y=-5/13-1/(22/169)(x-3)#
#y=-169/22x+6481/286#
#y=-7.6bar(81)x+22.66#
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Answer 2

To find the equation of the normal line of f(x) at x=3, we need to find the derivative of f(x) and evaluate it at x=3. Then, we can use the point-slope form of a line to determine the equation of the normal line.

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Answer 3

To find the equation of the normal line to the curve ( f(x) = \frac{(5-x)(4-x^2)}{x^3 - 1} ) at ( x = 3 ), we first need to find the derivative of ( f(x) ) and evaluate it at ( x = 3 ) to find the slope of the tangent line. Then, we'll use the negative reciprocal of this slope to find the slope of the normal line. Finally, we'll use the point-slope form of the equation of a line to find the equation of the normal line.

First, let's find the derivative of ( f(x) ):

[ f'(x) = \frac{d}{dx} \left( \frac{(5-x)(4-x^2)}{x^3 - 1} \right) ]

This requires using the quotient rule and the chain rule for differentiation. After finding ( f'(x) ), evaluate it at ( x = 3 ) to find the slope of the tangent line.

Next, once you have the slope of the tangent line, find the negative reciprocal of this slope to obtain the slope of the normal line.

Finally, knowing the slope of the normal line and the point of tangency ( (3, f(3)) ), you can use the point-slope form of the equation of a line:

[ y - y_1 = m(x - x_1) ]

where ( (x_1, y_1) ) is the point of tangency and ( m ) is the slope of the normal line. Substitute the values and simplify to obtain the equation of the normal line.

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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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