Find the equation of the tangent line to $f ( x ) = 4 \ln x ^ { 3 }$ at x = e.
A) $y = \frac { 12 } { e } x + 24$
B)y = 12x
C) $y = \frac { 12 } { e } x$
D) $y = \frac { 12 } { e } x - 24$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to $f(x) = 4\ln x^3$ at $x = e$, we first need to find the derivative of $f(x)$, which gives us the slope of the tangent line at any point $x$. The derivative of $f(x)$ using the chain rule is $f'(x) = 4 \cdot \frac{3}{x} = \frac{12}{x}$. Evaluating this derivative at $x = e$ gives us the slope of the tangent line at $x = e$: $f'(e) = \frac{12}{e}$.Next, we find the y-coordinate of the point on the curve at $x = e$ by substituting $x = e$ into the original function: $f(e) = 4\ln e^3 = 4 \cdot 3 = 12$.The equation of the tangent line can be written in point-slope form: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is a point on the line. Substituting $m = \frac{12}{e}$, $x_1 = e$, and $y_1 = 12$, we get:$y - 12 = \frac{12}{e}(x - e)$.To simplify, we distribute and rearrange to get the equation in slope-intercept form $y = mx + b$:$y = \frac{12}{e}x + b$.To find $b$, we use the fact that when $x = e$, $y = 12$:$12 = \frac{12}{e}e + b$, which simplifies to $12 = 12 + b$, but since we already used the point to derive the equation, we don't need to solve for $b$. The correct form of the equation is already found as:$y = \frac{12}{e}x$, which matches option C.

Find the Equation of the Tangent Line To f(x)=4ln⁡x3f ( x ) = 4 \ln x ^ { 3 }f(x)=4lnx3

Find the Equation of the Tangent Line To $f ( x ) = 4 \ln x ^ { 3 }$