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

User MJB
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1 Answer

2 votes

Answer: y = -4x + 4

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


User Ilia Ross
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5.3k points