iwillrektyou
iwillrektyou iwillrektyou
  • 23-10-2018
  • Mathematics
contestada

Find the equation of the tangent line to the graph f(x) = x(1-2x)^3 at the point (1,-1)

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tramserran
tramserran tramserran
  • 23-10-2018

Answer: y = -4x + 4

Step-by-step explanation:

use the product formula (ab' + a'b) to find the derivative:

x * (1 - 2x)³

a = x             a' = 1

b = (1 - 2x)³   b' = 3(-2)(1 - 2x)²

                         = -6(1 - 2x)²

 ab' + a'b

= x(-6)(1 - 2x)² + 1(x)

= -6x(1 - 2x)² + x

Plug in the given x-values to find the slope:

= -6(1)(1 - 2(1))² + (1)

= -6(-1)² + 1

= -6 + 1

= -5

Next, input the slope and the point into the Point-Slope formula:

y + 1 = -5(x - 1)

y + 1 = -4x + 5

     y = -4x + 4


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