Question

How do you solve the third part to this question? The question is attached as an image.

Answer 1

for my equation i got 4 + 1/16x but for some reason when i plug in 64.2 it won't work.

Answer 2

what do you get when you plugin x = 64.2 ?

Answer 3

should work
i love these historical questions, what life was like before the $3 calculator

Answer 4

using your equation I get
(64+ 0.2)/16 + 4 = 4 + 1/80 + 4 = 8.0125
if you square 8.0125 it should be close to 64.2

Answer 5

wait where did you get the .2 from?

Answer 6

your equation of a line is the equation of the tangent line to the curve y= sqr(x)
at x= 64
if you put in x=64 you will get out y=8 a perfect match to the curve.
as you move away from x=64, the y will be close (but not exactly) the same value as y= sqr(x)

they want you to use your linear equation to find y when x is 64.2

Answer 7

oh yes of course. i kept plugging in 64 by mistake

Answer 8

thanks

Answer 9

if you use x= 64 you get
y = x*1/16 + 4
y = 64/16 + 4 = 4 + 4 = 8

you are at the point of tangency where the line touches the curve.

Answer 10

i see

Answer 11

in the days before calculators (and even today), this "use the tangent line" rather than take the square root is much easier. It's a good trick, though obviously we have to know when the approximation is no longer "good enough" (i.e. as we move further away from the point of tangency)