`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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