"Linearization" is a fancy name for (and a particular use of) the tangent line.
(I do not have words to describe how much become clear when I realized what that sentence means.)
To approximate #sqrt79# by linearization, find the equation of a tangent line nearby. We want a tangent at a value of #x# near 79, where we can actually find #f(x)=sqrtx#.
Yes, there is an obvious choice we'll find the line tangent to the graph of #f(x)=sqrtx# at the point #(81,9)#.
We'e going to write the line in a particular way to emp[hasise the use we're about to make of it.
Point-slope form is: #y-y_1=m(x-x_1)# .
where the point #(x_1, y_1) # lies on the line and #m# = the slope of the line. (I'll put an additional note on this at the end.)
We'll use #y-f(81)= f'(81)(x-81)#.
With the tiniest bit of work, we can find #f'(81)=1/18#. so our line becomes:
#y-9= 1/18(x-81)#.
Now, if we wanted slope-intercept form, we would solve for #y#. But our intention is to use this line to approximate #f(x)# for #x# near #81#, so we start at #f(81)=9# and approximate the change in #y# by using the tangent line.
We write: #f(x)=9+1/18(x-81)# Or, using #L(x)# for the linearization:
#L(x)=9+1/18(x-81)#. Yes! Really! It's just the doggone tangent line!
#L(79)=9+1/18(79-81) = 9+1/18(-2)=9-2/18=9-1/9=9-0.1111=8.9999#.
Note:
A point #(x,y)# gets to be on the line through #(81,9)# with slope #m=1/18# by the following rule:
#(x,y)# is on this line if and only if the slope between #(x,y)# and #(81,9)# is #1/18#..
That is: if and only if #(y-9)/(x-81) = 1/18# Clear the fraction and you've got point-slope form.
