Given

f(x) = √(x-12) \ \

Δx = x-x₀ = 0.4 \ \

f(37) = √(37-12) = √25 = 5 \ \

f(37.4) = √(37.4-12) = √25.4 \ \

f(x) = √(x-12) \ \

Apply diffenceation. \ \

f \\'(x) = 1/[2√(x-12)] \ \

Substitute x = 37 \ \

f \\'(37) = 1/[2√(37-12)] \ \

f \\'(37) = 1/[2√(25)] \ \

f \\'(37) = 1/(2*5) \ \

f \\'(37) = 1/10 \ \

Formula for Linear Approximation : f(x) = f(x₀) + (x-x₀)f \\'(x₀) \ \

Consider x = 37.4 and  x₀ = 37 \ \

f(37.4) = f(37) + (0.4)f \\'(37) \ \

Substitute f \\'(37) = 1/10 \ \

f(37.4) = f(37) + (0.4)(1/10) \ \

f(37.4) - f(37) = 0.04 \ \

The solution is f(37.4) - f(37) = 0.04 \ \