f(x)=2x^3+x^2-4 and h = 0.5, estimate f^'(2.5) and f^('')(2.5)
using Two point Forward difference, Backward difference, Central difference formula numerical differentiation
Also find exact value of f', f'' and error for each estimation
Solution:
Equation is f(x) = 2x^3+x^2-4.
:. f^'(x) = 6x^2+2x
:. f^('')(x) = 12x+2
The value of table for x and y
Two-point FDF (Forward difference formula)
f^'(x)=(f(x+h)-f(x))/h
f^'(2.5)=(f(2.5+0.5)-f(2.5))/0.5
f^'(2.5)=(f(3)-f(2.5))/0.5
f^'(2.5)=(59-33.5)/0.5
f^'(2.5)=51
Absolute Error:|"exact value of " f^'(2.5)-(51)|=|42.5 -51|=8.5
Two-point BDF (Backward difference formula)
f^'(x)=(f(x)-f(x-h))/h
f^'(2.5)=(f(2.5)-f(2.5-0.5))/0.5
f^'(2.5)=(f(2.5)-f(2))/0.5
f^'(2.5)=(33.5-16)/0.5
f^'(2.5)=35
Absolute Error:|"exact value of " f^'(2.5)-(35)|=|42.5 -35|=7.5
Two-point CDF (Central difference formula)
f^'(x)=(f(x+h)-f(x-h))/(2h)
f^'(2.5)=(f(2.5+0.5)-f(2.5-0.5))/(2*0.5)
f^'(2.5)=(f(3)-f(2))/1
f^'(2.5)=(59-16)/1
f^'(2.5)=43
Absolute Error:|"exact value of " f^'(2.5)-(43)|=|42.5 -43|=0.5
This material is intended as a summary. Use your textbook for detail explanation.
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