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

Answer 1

Emilia Aguilar

#y=7x-9#

#y=kx+n#, where #k=f'(x_0), x_0=1#

Let's use the quotient rule to determine the first derivative:

#f'(x)=(4(3-4x^2-x)-4x(-8x-1))/(3-4x^2-x)^2#

#f'(x)=(12-16x^2-4x+32x^2+4x)/(3-4x^2-x)^2#

#f'(x)=(16x^2+12)/(3-4x^2-x)^2#

#k=f'(x_0)=f'(1)=(16+12)/(3-4-1)^2=28/(-2)^2=7#

For #x_0=1 => y_0=f(1)=4/(3-4-1)=4/(-2)=-2#

#y_0=kx_0+n => -2=7*1+n => n=-9#

Finally,

#y=7x-9#

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Answer 2

Scarlett Allison

To find the equation of the tangent line of f(x) at x = 1, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the quotient rule:

f'(x) = [(3-4x^2-x)(4) - (4x)(-8x-1)] / (3-4x^2-x)^2

Next, we substitute x = 1 into f'(x) to find the slope of the tangent line at x = 1:

f'(1) = [(3-4(1)^2-1)(4) - (4(1))(-8(1)-1)] / (3-4(1)^2-1)^2

Finally, we substitute the slope and the point (1, f(1)) into the point-slope form of a line to find the equation of the tangent line.

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Answer from HIX Tutor

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