Question

I really don't know where to start or go with this question...There aren't any similar examples in my book so if you could walk me through it so I can be prepared for my test - that would be great! Thanks.

(a) Approximate √26 by using the linearization of y=√x at the point (25,5). Show the computation that leads to your conclusion.

(b) Approximate √26 by using a first guess of 5 and one iteration of Newton's method to approximate the zero of x^2-26. Show the computation that leads to your conclusion.

(c) Approximate ³√26 by using an appropriate linearization. Show the computation that leads to your conclusion.

Answer 1

y = m x + b

at (25,5)

5 = m (25) + b
what is m, the slope, there?
y = x^.5
dy/dx = .5 x^-.5 = .5/5 = .1 = m
so
5 = .1 (25) + b
2.5 = b
so our line through (25,5) is
y = .1 x + 2.5
now if x = 26
y = 2.6 + 2.5 = 5.1
that is part a
check it
sqrt 26 = 5.099 Awful close !!! :)

Answer 2

b should come out just like a :)

Answer 3

Thanks so much!! How would I go about C, if you get the chance? A & B make sense now thanks to that walkthrough...I'm just unsure of how to go about it with the cubed root.

Answer 4

Approximate ³√26 by using an appropriate linearization.

y = x^(1/3)

we know that if x = 27, y = 3 so I will start there. Of course you could do something else.

dy/dx = (1/3) x^-2/3
at x = 27
dy/dx = (1/3) /9 = 1/27 = m
so
y = (1/27) x + b
3 = (1/27)(27) + b
b = 2
so
y = (1/27) x + 2
if x = 26
y = 2.963

now for real to check
26^(1/3) = 2.962 We did good again :)

Answer 5

oops! disregard my previous post. thanks so much, I truly appreciate it. Gonna ace this next test :)

Answer 6

Good luck :)