Question

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

• A. $-34.995$
• B. $-33.995$
• C. $-33.335$
• D. $-35.993$

Answer 1

The correct option is A $-34.995$

Firstly break the number $5.001$ as $x=5$ and $△x=0.001$ and use the relation $f(x+△x)=f\left(x\right)+△x{f}^{\prime }\left(x\right)$.
Consider, $f\left(x\right)=x^3-7x^2+15$
$\Rightarrow f^{\prime }\left(x\right)=3x^2-14x$
Let $x=5$
and $△x=0.001$
Also, $f(x+△x)=f\left(x\right)+△x{f}^{\prime }\left(x\right)$
Therefore, $f(x+△x)=(x^3-7x^2+15)+△x(3x^2-14x)$
$\Rightarrow f\left(5.001\right)=(5^3-7\times 5^2+15)+(3\times 5^2-14\times 5)\left(0.001\right)$
(as $x=5,△x=0.001)$
$=125-175+15+(75-70)\left(0.001\right)$
$=035+\left(5\right)\left(0.001\right)$
$=-35+0.005=-34.995$