Question

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

• A. $34.995$
• B. $-30.995$
• C. $24.875$
• D. None of these

Answer 1

Answer: (A)

Consider $f(x)=x^3-7x^2+15$
$\Rightarrow f^{\prime }(x)=3x^2-14x$
Let $x=5$ and $\Delta x=0.001$
Also, $f(x+\Delta x)\approx f(x)+\Delta x{f}^{\prime }(x)$
Therefore, $f(x+\Delta x)=\left(x^3-7x^2+15\right)+\Delta x\left(3x^2-14x\right)$
$\Rightarrow f(5.001)=\left(5^3-7\times 5^2+15\right)+\left(3\times 5^2-14\times 5\right)(0.001)$
$($ as $x=5,\Delta x=0.001)$
$=125-175+15+(75-70)(0.001)$
$=-34.995$