"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√79 by linearization, find the equation of a tangent line nearby. We want a tangent at a value of xx near 79, where we can actually find f(x)=sqrtxf(x)=√x.
Yes, there is an obvious choice we'll find the line tangent to the graph of f(x)=sqrtxf(x)=√x at the point (81,9)(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)y−y1=m(x−x1) .
where the point (x_1, y_1) (x1,y1) lies on the line and mm = 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.
