Q.

The approximate value of f(5.001), where f(x)=x3−7x2+15.

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a

-34.995

b

34.995

c

-38.458

d

38.458

answer is A.

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Detailed Solution

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
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