Find an equation of the tangent line to the graph of y = $x ^ { 3 }$ ln(-2x) at x = -1.
A) y = (x + ln 2) - 1
B) y = (1 + 3 ln 2)(x + 1) - ln 2
C) y - 1 = 4(x - ln 2)
D) y + ln 2 = 4(x - 1)

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the graph of $y = x^3 \ln(-2x)$ at $x = -1$, we first need to find the derivative of $y$ with respect to $x$, which will give us the slope of the tangent line at any point $x$. The derivative, using the product rule and chain rule, is:$$y' = 3x^2 \ln(-2x) + x^3 \cdot \frac{1}{-2x} \cdot (-2) = 3x^2 \ln(-2x) - x^2$$At $x = -1$, the slope of the tangent line ($m$) is:$$m = 3(-1)^2 \ln(-2(-1)) - (-1)^2 = 3 \ln(2) - 1$$The $y$-coordinate when $x = -1$ is:$$y = (-1)^3 \ln(-2(-1)) = -\ln(2)$$Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line (in this case, $(-1, -\ln(2))$), we get:$$y + \ln(2) = (3 \ln(2) - 1)(x + 1)$$This simplifies to the equation given in choice B.

Find an Equation of the Tangent Line to the Graph x3x ^ { 3 }x3

Find an Equation of the Tangent Line to the Graph $x ^ { 3 }$