Answer:
Step-by-step explanation:
You could do this 2 ways (NOT including going to a calculator and punching it in and getting the answer!). The first way is to use Newton's Method (which will give you an EXACT answer); the second is to use differentials (which will only approximate the answer). We'll do both here, beginning with Newton's Method. The formula for Newton's Method is
[tex]x_{1}-\frac{f(x_{1})}{f'(x_{1})}[/tex] where x1 is your initial input.
Let x = [tex]\sqrt{38}[/tex] so
[tex]x^2=38\\x^2-38=0[/tex] with a derivative of 2x. Fitting this into the formula with an initial input of 6, since 6-squared is 36 and that's pretty close.
[tex]6-\frac{6^2-38}{2(6)}=\frac{37}{6}[/tex] and use that answer in the next iteration:
[tex]\frac{37}{6}-\frac{\frac{37}{6}^2-38 }{2(\frac{37}{6}) }=\frac{2737}{444}[/tex] and use that answer in the next iteration:
[tex]\frac{2737}{444}-\frac{\frac{2737}{444}^2-38 }{2(\frac{2737}{444}) }=6.164414003[/tex] and use that answer in the next iteration:
[tex]6.164414003-\frac{.164414003^2-38}{2(6.164414003)}=6.164414003[/tex]
Since we got the same number after that last iteration, the exact value of the square root of 38 is 6.164414003. Check your calculator for accuracy.
The next method is by differentials. the formula is
f(x + Δx) ≈ f(x) + f'(x)dx
We will let f(x) = √x with x = 36 and dx = 2 (36 is a perfect square and adding 2, the dx value, gets you back to 38, our number in question).
Filling in the formula:
f(x + Δx) ≈ √x + [tex]\frac{1}{2\sqrt{x} }[/tex] dx
f(x + Δx) ≈ [tex]\sqrt{36}+\frac{1}{2\sqrt{36} }(2)[/tex] and simplifying a bit:
f(x + Δx) ≈ 6 + [tex]\frac{1}{6}[/tex] which gives us 6.166666
Not quite as accurate as Newton's Method, but it works for mere approximations.
