Find the equation of the tangent line to the curve when x has the given value.
-$$f ( x ) = \frac { 6 } { x + 3 } ; x = 2$$
A) $y = \frac { 6 } { 25 } x + \frac { 18 } { 25 }$
B) $y = - \frac { 6 } { 25 } x + \frac { 42 } { 25 }$
C) $y = - \frac { 6 } { 25 } x + \frac { 18 } { 25 }$
D) $y = - \frac { 12 } { 25 } x + \frac { 42 } { 25 }$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line, 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 $f(x) = \frac{6}{x + 3}$ is $f'(x) = -\frac{6}{(x + 3)^2}$. Substituting $x = 2$ into $f'(x)$ gives us the slope $m = -\frac{6}{(2 + 3)^2} = -\frac{6}{25}$. The value of the function at $x = 2$ is $f(2) = \frac{6}{2 + 3} = \frac{6}{5}$. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with $x_1 = 2$ and $y_1 = \frac{6}{5}$, we get $y - \frac{6}{5} = -\frac{6}{25}(x - 2)$. Simplifying this equation gives us $y = -\frac{6}{25}x + \frac{42}{25}$, which matches choice B.

Find the Equation of the Tangent Line to the Curve f(x)=6x+3;x=2f ( x ) = \frac { 6 } { x + 3 } ; x = 2f(x)=x+36​;x=2

Find the Equation of the Tangent Line to the Curve $f ( x ) = \frac { 6 } { x + 3 } ; x = 2$