An equation for the tangent line to the curve $y = \left( x ^ { 3 } + x - 1 \right) ^ { 3 }$ at the point where x = 1 is
A) y = 12x - 11
B) y = 12x
C) y = 3x + 1
D) y = 12x - 1

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the curve at a specific point, we first need to find the derivative of the given function to determine the slope of the tangent line at that point. The given function is $y = (x^3 + x - 1)^3$. Using the chain rule, the derivative $y'$ is $3(x^3 + x - 1)^2 \cdot (3x^2 + 1)$. Substituting $x = 1$ into the derivative gives $y'(1) = 3(1^3 + 1 - 1)^2 \cdot (3(1)^2 + 1) = 3 \cdot 1^2 \cdot 4 = 12$, which is the slope of the tangent line at $x = 1$. The y-value at $x = 1$ is $y = (1^3 + 1 - 1)^3 = 1$. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m = 12$, $x_1 = 1$, and $y_1 = 1$, we get $y - 1 = 12(x - 1)$, which simplifies to $y = 12x - 11$.

An Equation for the Tangent Line to the Curve y=(x3+x−1)3y = \left( x ^ { 3 } + x - 1 \right) ^ { 3 }y=(x3+x−1)3

An Equation for the Tangent Line to the Curve $y = \left( x ^ { 3 } + x - 1 \right) ^ { 3 }$