Question

The approximate value of $\sqrt[3]{3-0.99}$ is

• A. $-0.9967$
• B. $-0.9976$
• C. $-1.0033$
• D. $-1$

Answer 1

The correct option is A $-0.9967$

Here, $f\left(x\right)=\sqrt[3]{3x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{3}\frac{1}{\sqrt[3]{3x^2}}$

For $-0.99$, $x=-1$ and $\Delta x=0.01$
Now, $f(x+\Delta x)=f\left(x\right)+f^{\prime }\left(x\right)\Delta x$
$f\left(\sqrt[3]{3-0.99}\right)=f(-1)+\left(f^{\prime }\right(-1\left)\right)\left(0.01\right)$
$\Rightarrow f\left(\sqrt[3]{3-0.99}\right)=-1+\frac{1}{3}\frac{1}{\sqrt[3]{3(-1{)}^2}}\left(0.01\right)$
$\Rightarrow f\left(\sqrt[3]{3-0.99}\right)=-1+\frac{1}{3}\left(0.01\right)$
$\Rightarrow f\left(\sqrt[3]{3-0.99}\right)=-0.9967$