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-(96)/(x^3) \end{gathered}`

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

`\begin{gathered} y^(\prime)(2)=3-(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)+(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