Write down the parabola equation. A top open parabola's equation has the form:
Y= f(x) = aX^2 +bX +c = F(x)
where:
a, b, c are numeric constants
Y, X are the variables,
Remember, the aX^2 term is always positive.
For example, assume:
Y = f(x) = X^2
a = 1, b = 0, c = 0
And we wish to find Y = (3.02) = (3.02)^2
Write down the formula for the linear approximation. The formula is:
f(X) = f(Xo) + ( f'(Xo) (X - Xo) )
where:
f(X) is the unknown value
f(Xo) is the know value
f'(Xo) is the derivative for the input
Xo is the input for the known value
X is the input for the value to be found
From the example, 3.02 is very close to 3 (which is simple to calculate: 3^2=9), therefore we have:
X = 3.02
Xo = 3
f'(Xo) = 2X
f(X) = f(Xo) + [ f'(Xo) (X - Xo) ]
f(3.02) = f(3) + [ f'(3) (3.02 -3) ]
Find the derivative of the open parabola equation. Replace the derivative on the equation.
f(X) = X^2
f'(X) = 2X
f'(3) = (2)(3) = 6
Replace the derivative in the formula for linear approximation. Solve the formula and find the answer.
f(X) = f(Xo) + [ f'(Xo) (X - Xo) ]
f(3.02) = f(3) + [ (6) (3.02 -3) ]
f(3.02) = 9 + [ (6) ( 0.02) ]
f(3.02) = 9 + 0.12
f(3.02) = 9.12
Using a calculator (3.02)^2 = 9.1204, which checks that the linearization is a fast and accurate tool.