linear approximation

[dans]

approximate 31.4^2/5

 

[Gene]

f(x) = x^(2/5)
f'(x) = (2/5)x^(-3/5)
x₀ = 32
f(x₀) = 4
f'(x₀) = (2/5)/32^(3/5) = 1/20
x = 31.4
f(x) = f(x₀) + f'(x₀)(x-x₀)

 

[elchichita]

f(x) = x^(2/5)
f'(x) = (2/5)x^(-3/5)
x₀ = 32
f(x₀) = 4
f'(x₀) = (2/5)/32^(3/5) = 1/20
x = 31.4
f(x) = f(x₀) + f'(x₀)(x-x₀)

Correct...You also coul take x₀ = 31, and proceed similary

 

[dans]

Thank you. I can understand the problem and solution now.

 

[tkhunny]

approximate 31.4^2/5

Using differential notation:

f(x) = x^(2/5)
df(x,dx) = (2/5)*[x^(-3/5)]*dx
dx = 31.4 - 32 = -0.6 = -3/5

f(32) = 32^(2/5) = 2^2 = 4
df(x,dx) = (2/5)*[32^(-3/5)]*(-3/5) = (2/5)*(-3/5)*(2^(-3)) = (-6/25)*(1/8) = -3/100

f(31.4) =~ f(32) + df(32,-3/5) = 4 - 3/100 = 3.97

How good is it? 31.4^(2/5) = 3.96982954... Not bad.