The approximate value of f(5.001), where f(x)=x3−7x2+15.
-34.995
34.995
-38.458
38.458
Let x=5 and Δx=0.001. Then, we have
f(5.001)=f(x+Δx)=(x+Δx)3−7(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)3−7(5)2+15+3(5)2−14(5)(0.001)=(125−175+15)+(75−70)(0.001)=−35+0.005=−34.995