The approximate value

(a) 2.0125

Consider the function $y=f(x)=x^{\frac{1}{5}}$.

Let :

$x=32$

$x+\Delta x=33$

$\Rightarrow \Delta x=1$

$y=(x)^{\frac{1}{5}}$

For $x=32$,

$y=2$

$\mathrm{AlsO}, \frac{d y}{d x}=\frac{1}{5(x)^{\frac{4}{5}}}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=32}=\frac{1}{80}$

$\Rightarrow \Delta y=d y=\frac{d y}{d x} d x=\frac{1}{80} \times 1=0.0125$

$\therefore(33)^{\frac{1}{5}}=y+\Delta y=2.0125$