Suppose fxex3x3 Approximate the derivative fx for x110511011

Suppose f(x)=e^x/3+x³. Approximate the derivative f(x) for x=1,1.05,1.10,1.15 using the function values at these points. Apply the Three-point Midpoint method where applicable, and three-point end-point otherwise. Also estimate the error of the approximation.

Solution

1st we will find three points end points we get

Using the 3-point endpoint formula with h = 0.1 gives

1/0.2[3f (2.0) + 4f (2.1) f (2.2]
= 5[3(14.778112)+ 4(17.148957) 19.855030)]

22.032310
and with h = 0.1 gives 22.054525.
Using the 3-point midpoint formula with h = 0.1 gives


1/0.2[f (2.1) f (1.9] now ,

so, The only five-point formula for which the table gives sufficient data is
the midpoint formula with h = 0.1. This gives

1/0.2[f (1.8) 8f (1.9) + 8f (2.1) f (2.2)]
=

1/0.2[10.889365 8(12.703199)+ 8(17.148957) 19.855030]
=22.166999

The true value in this case is f (2.0) = 1/3e^x/3+3x^2 = 22.167168, so the
approximation errors are actually:
Method h Approximation Error
Three-point endpoint 0.1 1.35 × 101
Three-point endpoint 0.1 1.13 × 101
Three-point midpoint 0.2 2.47 × 101
Three-point midpoint 0.1 6.16 × 102
Five-point midpoint 0.1 1.69 × 104