How do you find the tangent line approximation to #f(x)=1/x# near #x=1# ?

Question

Cameron Alexander · Accepted Answer

#L(x) = 2-x#

We find the answer from the linear approximation #L(x) = f(a)+f'(a)(x-a)#. For linear approximations, we want both #f(a)# and #f'(a)# so that they are easy to calculate. In this case, #f(1)# and #f'(1)# are easy.

To find the derivative of

#f(x) = 1/x#

it is best to rewrite it in power form:

#f(x) = x^(-1)#

Using the power rule for derivatives:

#f'(x)=-x^(-2)#

So substituting, #a=1#:

#f(a)=1/1#

#f'(1)=-1/1^2#

Substituting this into #L(x)#,

#L(x)=1-1(x-1)#

#=1-x+1#

#=2-x#

Ezekiel Alexander · Answer

undefined To find the tangent line approximation to $ f(x) = \frac{1}{x} $ near $ x = 1 $, we use the formula for the equation of a tangent line:

$$ y = f'(a)(x - a) + f(a) $$

where $ f'(a) $ is the derivative of $ f(x) $ evaluated at $ x = a $, and $ f(a) $ is the value of $ f(x) $ at $ x = a $.

First, we find the derivative of $ f(x) = \frac{1}{x} $:
$$ f'(x) = -\frac{1}{x^2} $$

Then, we evaluate $ f'(x) $ at $ x = 1 $:
$$ f'(1) = -\frac{1}{1^2} = -1 $$

Next, we find the value of $ f(x) $ at $ x = 1 $:
$$ f(1) = \frac{1}{1} = 1 $$

Now, we have the slope of the tangent line ($ f'(1) = -1 $) and the point ($ 1, 1 $) through which the tangent line passes. Using these values, we can write the equation of the tangent line:
$$ y = -1(x - 1) + 1 $$

$$ y = -x + 1 + 1 $$

$$ y = -x + 2 $$

So, the tangent line approximation to $ f(x) = \frac{1}{x} $ near $ x = 1 $ is $ y = -x + 2 $.