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

Question

Quizplus · Accepted Answer

To find the equation of the tangent line at $x = 1$, we first need to calculate the derivative of $y = 2x + \ln\left(\frac{1}{x}\right)$ to find the slope of the tangent line at that point. The derivative of $y$ with respect to $x$ is $y' = 2 - \frac{1}{x}$. Substituting $x = 1$ into the derivative gives $y' = 2 - 1 = 1$, which is the slope of the tangent line at $x = 1$. The value of $y$ at $x = 1$ is $y = 2(1) + \ln(1) = 2$. Therefore, the equation of the tangent line using the point-slope form is $y - y_1 = m(x - x_1)$, where $m = 1$, $x_1 = 1$, and $y_1 = 2$. Substituting these values gives $y - 2 = 1(x - 1)$, which simplifies to $y = x + 1$.

Find an Equation of the Tangent Line to the Graph (1x)\left( \frac { 1 } { x } \right)(x1​)

Find an Equation of the Tangent Line to the Graph $\left( \frac { 1 } { x } \right)$