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:

$f^{\prime}(x)=(\frac{1}{x})^{\prime}=(x^{-1})^{\prime}=-1x^{-1-1}=-\frac{1}{x^2}\text{.}$

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

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

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

$y-f(-7)=-\frac{1}{49}(x-(-7))\text{.}$

Simplifying the above result we get:

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

Subtracting 1/7 from the above equation we get:

$\begin{gathered} y+\frac{1}{7}-\frac{1}{7}=-\frac{1}{49}x-\frac{1}{7}-\frac{1}{7}, \\ y=-\frac{1}{49}x-\frac{2}{7}\text{.} \end{gathered}$

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:

$-\frac{1}{-\frac{1}{49}}=49$

Using the slope-point formula we get that:

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

Substracting 1/7 from the above equation we get:

$\begin{gathered} y+\frac{1}{7}-\frac{1}{7}=49x+343-\frac{1}{7}, \\ y=49x+\frac{2400}{7}\text{.} \end{gathered}$

Answer:

(a)

$y=-\frac{1}{49}x-\frac{2}{7}\text{.}$

(b)

$y=49x+\frac{2400}{7}\text{.}$