Question

Find the equation of the tangent to the curve at the point where x = 2, givingthe answer in the formy= mx + C.

Answer 1

First, find the derivative of the curve:

$\begin{gathered} y^{\prime}=3(1)+48(-2\cdot x^{-3}) \\ y^{\prime}=3-\frac{96}{x^3} \end{gathered}$

Now, evaluate the point into the derivative to find the slope:

$\begin{gathered} y^{\prime}(2)=3-\frac{96}{2^3} \\ y^{\prime}(2)=m=3-12=-9 \end{gathered}$

Find the coordinate y for the point x =2 evaluating x = 2 into the function:

$\begin{gathered} y(2)=3(2)+\frac{48}{2^2} \\ y(2)=6+12=18 \end{gathered}$

Use the point-slope equation:

$\begin{gathered} y-y1=m(x-x1) \\ y-18=-9(x-2) \\ y-18=-9x+18 \\ y=-9x+36 \end{gathered}$

Answer:

y = -9x + 36