Chapter 1.3, Problem 40E

In Exercises 33-40, find an equation of the tangent line to the graph $y=f(x)$

at the given $x$. Do not apply formula (6), but proceed as we did in Example 4.

$f(x)=\frac{1}{x^3},x=3$

Example 4

Finding the Equation of Tangent Line at Given $x$

Find the point-slope equation of the tangent line to the graph of $f(x)=\frac{1}{x^2}$

at $x=2$.

Solution

In this problem, we are given a point on the tangent line, but only its first coordinate $x=2$.

Since the point is on the graph $f(x)=\frac{1}{x^2}$, we get the second coordinate by plugging $x$

-value into $f(x)$:

$y=f(2)=\frac{1}{2^2}=\frac{1}{4}$.

Thus, $\left(2,\frac{1}{4}\right)$

is the point on the graph and the tangent line. Next, we find the slope of the tangent line. For this purpose, we compute $f\text{'}(x)$

by using the power rule:

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

$f\text{'}(x)=(-2)x^{-2-1}=-2x^{-3}=\frac{-2}{x^3}$.

The slope tangent line when $x=2$

is $f\text{'}(2)=\frac{-2}{2^3}=\frac{-1}{4}$.

In the point-slope form, the equation of the tangent line is $y-\frac{1}{4}=-\frac{1}{4}(x-2)$