First slide
Differentials; errors and approximations
Question

The approximate value of f(5.001), where f(x)=x37x2+15

Moderate
Solution

 Let x=5 and Δx=0.001. Then, we have 

f(5.001)=f(x+Δx)=(x+Δx)37(x+Δx)2+15 Now, Δy=f(x+Δx)f(x)

f(x+Δx)=f(x)+Δy Or  f(x)+f(x)Δx ( as dx=Δx) Or f(5.001)=(5)37(5)2+15+3(5)214(5)(0.001)=(125175+15)+(7570)(0.001)=35+0.005=34.995

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