Let be a real number. Let x~2.5 be an approximate value of with absolute error
at most 0.01. The function ( )
is evaluated at instead of . Estimate the absolute error
for example f(x)=x2f(x)=x^2f(x)=x2
then absolute error of f(x) at x = 2.5:
∣f(x+Δx)−f(x)∣=f(2.51)−f(2.5)=2.512−2.52=0.0501|f(x+\Delta x)-f(x)|=f(2.51)-f(2.5)=2.51^2-2.5^2=0.0501∣f(x+Δx)−f(x)∣=f(2.51)−f(2.5)=2.512−2.52=0.0501
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