Firstly, break the number 5.001 as x=5 and Δx=0.001 and use the relation f(x+Δx)≃f(x)+Δxf′(x) Consider, f(x)=x3−7x2+15 ⇒ f′(x)=3x2−14x Let x = 5 and Δx=0.001 Also, f(x+Δx)≃f(x)+Δxf′(x) Therefore, f(x+Δx)=(x3−7x2+15)+Δx(3x2−14x) ⇒ f(5.001)≃(53−7×52+15)+(3×52−14×5)(0.001) (as x=5,Δx=0.001) =125−175+15+(75−70)(0.001) =−35+(5)(0.001) =−35+0.005=−34.995