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Manipal MET2014MathematicsApplication of Derivatives

The approximate value of f(5.001) , where f(x)=x^3-7 x^2+15 , is

Options

  1. A-34.995
  2. B-33.995
  3. C-33.335
  4. D-35.993

Correct answer

A. -34.995

Step-by-step solution

Firstly, break the number 5.001 as x=5 and x=0.001 and use the relation f(x+ x)=f(x)+ x f^ (x) . Consider, f(x)=x^3-7 x^2+15 f^ (x)=3 x^2-14 x Let x=5 and x=0.001 Also, f(x+ x) f(x)+ x f^ (x) Therefore, aligned & f(x+ x)= (x^3-7 x^2+15 )+ x (3 x^2-14 x ) & f(5.001) (5^3-7 5^2+15 ) & + (3 5^2-14 5 )(0.001) aligned aligned ( as x & =5, x=0.001) & =125-175+15+(75-70)(0.001) & =-35+(5)(0.001) & =-35+0.005=-34.995 aligned

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