How do you find the square root of 79 using linearization techniques?

Question

Charles Arocha · Accepted Answer

"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.

Charles Anctil · Answer

undefined To find the square root of 79 using linearization techniques, we start by choosing a number close to the square root of 79. Let's choose 8, as $8^2 = 64$, which is close to 79. Then, we calculate the derivative of the function $f(x) = \sqrt{x}$ at $x = 64$, which is $f'(x) = \frac{1}{2\sqrt{x}}$.

Next, we use the linear approximation formula:

$$f(x) \approx f(a) + f'(a)(x - a)$$

where $a$ is the number we chose (in this case, 8). Substituting the values, we get:

$$f(x) \approx \sqrt{64} + \frac{1}{2\sqrt{64}}(x - 64)$$

Simplifying, we have:

$$f(x) \approx 8 + \frac{1}{16}(x - 64)$$

Now, we substitute $x = 79$ into the equation to approximate the square root of 79:

$$f(79) \approx 8 + \frac{1}{16}(79 - 64)$$

$$f(79) \approx 8 + \frac{1}{16}(15)$$

$$f(79) \approx 8 + \frac{15}{16}$$

$$f(79) \approx 8.9375$$

So, the square root of 79, using linearization techniques, is approximately 8.9375.