What is the v43 ?

Question

√43=6+(43-36)/(2√36)

=6+7/12

=6+5833

=6.68

Is this correct

 

 

Answer 1

Cosaque, the process you used is correct but you added wrong, 6 + 0.5833 = 6.5833 not 6.658.  This is a calculus problem and are usually worded in this way

 

Use a linear approximation (or differential) to estimate the given number √43.

 

You used the following

 

f(x + Δx) = f(x) +dy = f(x) + f '(x)dx,  where x = 36, dx = Δx = 7, f(x) = √x, f '(x) = 1/(2√x)

 

f(36 + 7) = sqrt(36) + (1/(2sqrt(36))(7) = 6 + 7/12 ≈ 6.58

 

 

Answer 2

Let f(x) = √x  So, f'(x) = 1/(2√x)

 

If x is close to a, then f(x) ≈ f(a) + f'(a)(x-a)

 

When a = 36, we have f(x) ≈ f(36) + f'(36)(x-36)

 

                                      = √36 + 1/(2√36)(x-36)

 

So, using the formula with x = 43, we would have:

 

√43 ≈ 6 + (1/12)(7) = 6.5833333

 

However, 43 is not close to 36, so the "approximation" isn't all that great.  

 

√43 ≈ 6.5574385

 

              

Answer 3

I don't quite understand what you are asking. Are you trying to find the approximated value of sqrt(43)?

Answer 4

From where did you get the process you used?

By calculator, √43 ≈ 6.557438524