Question

Use a linear approximation (or differentials) to estimate the given number.

sqrt(99.6)

so i think i start with f(x)=sqrt(100-x)
so f'(x)= 1/2(100-x)^-1/2 (-1) using a=0 f(x)= 10 and f'(x)= -0.05

when using L(x)=f(a)+f'(a)(x-a) where x=a I get 10 + -0.05(x) and then i plug in 99.6 for x and get 5.02 which is definitely not correct.

Answer 1

y = √x
y^2 = x
2y dy = dx
dy = dx/(2y)
So, given
x = 100
y = 10
dx = -0.4

dy = -0.4/20 = -0.02
so, √(99.6) ≈ 10-.02 = 9.98
actual value: 9.97998

As you can see, rather than using √(100-x) you just use √x and make a small adjustment. That way you are using dx and dy to approximate the value, rather than just plugging in a new value for x.

Answer 2

okay , thanks!

Answer 3

For webassign its 9.98.