Question

1. Let f(x): 1/xa) Find the tangent line to the graph of the function at x =-7.y=____b) Find the perpendicular line to the graph of the function at x = -7. Using the slope from the part a, the perpendicular slope is the opposite reciprocal of that slope valuey= ____

Answer 1

To answer this question we will use the following slope-point formula:

`y-y_1=m(x-x_1)\text{.}`

Recall that the slope of the tangent line to the graph of a function h(x) at (x,h(x)) is h'(x).

Now, we know that:

`f^(\prime)(x)=((1)/(x))^(\prime)=(x^(-1))^(\prime)=-1x^(-1-1)=-(1)/(x^2)\text{.}`

a) Evaluating f'(x) at x=-7 we get:

`f^(\prime)(-7)=-(1)/((-7)^2)=-(1)/(49)\text{.}`

Using the slope-point formula we get that the tangent line to the graph of the function at x=-7 is:

`y-f(-7)=-(1)/(49)(x-(-7))\text{.}`

Simplifying the above result we get:

`\begin{gathered} y-(1)/(-7)=-(1)/(49)(x+7), \\ y+(1)/(7)=-(1)/(49)x-(1)/(7)\text{.} \end{gathered}`

Subtracting 1/7 from the above equation we get:

`\begin{gathered} y+(1)/(7)-(1)/(7)=-(1)/(49)x-(1)/(7)-(1)/(7), \\ y=-(1)/(49)x-(2)/(7)\text{.} \end{gathered}`

b) Now, recall that the product of the slopes of two perpendicular lines is -1, therefore, the slope of the perpendicular line to the tangent line of the given function at x=-7 is:

`-(1)/(-(1)/(49))=49`

Using the slope-point formula we get that:

`\begin{gathered} y-f(-7)=49(x-(-7)), \\ y+(1)/(7)=49x+343. \end{gathered}`

Substracting 1/7 from the above equation we get:

`\begin{gathered} y+(1)/(7)-(1)/(7)=49x+343-(1)/(7), \\ y=49x+(2400)/(7)\text{.} \end{gathered}`

Answer:

(a)

`y=-(1)/(49)x-(2)/(7)\text{.}`

(b)

`y=49x+(2400)/(7)\text{.}`