Question

If $f\left(x\right)=x^3-7x^2+15,$then the approximate value of f(5.001) is

• A. - 34.995
• B. 34.995
• C. 35.995
• D. -35.995

Answer 1

The correct option is A

- 34.995

$x=5,8x=0.001f\left(x\right)=x^3-7x^2+15f\left(5\right)=125-175+5=-35f^{\prime }\left(x\right)=3x^2-14x{f}^{\prime }\left(5\right)=75-70f^{\prime }\left(5.001\right)\simeq f\left(5\right)+f^{\prime }\left(5\right)\left(0.001\right)=-35+0.005=-34.995$