COMEDK2023MathematicsApplication of Derivatives

The approximate value of f(5.001) , where f(x)=x^3-7 x^2+10

Options

A-39.995
B-38.995
C-37.335
D-40.995

Correct answer

A. -39.995

Step-by-step solution

First, break the number 5.001 as x=5 and x=0.001 and use the relation aligned & f(x+ x) f(x)+ x f^ (x) & consider f(x)=x^3-7 x^2+10 f^ (x)=3 x^2-14 x aligned Therefore, aligned & f(x+ x) (x^3-7 x^2+10 )+ x (3 x^2-14 x ) & f(5.001) (5^3-7(5)^2+10 )+(0.001) & (3(5)^2-14(5) ) & =(125-175+10)+(0.001)(75-70) & =-40+(0.001)(5)=-40+0.005=-39.995 aligned

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