What is the value of $\sqrt[3]{26}$ corrected up to three decimal places?
What is the value of $\sqrt[3]{26}$ corrected up to three decimal places?
- 2.998
- 2.844
- 2.962
- 2.823
Solution
Let $y=f(x)=x^{1 / 3}$
$
\begin{aligned}
x+\Delta x & =26 \\
27+\Delta x & =26 \quad[\because \text { let } x=27]
\end{aligned}
$
$
\Delta x=-1
$
We have, $\quad y=x^{1 / 3}$
Differentiate w.r.to ' $x$ '
$
\begin{aligned}
\frac{d y}{d x} & =\frac{1}{3} x^{\frac{1}{3}-1} \\
d y & =\frac{1}{3} x^{-\frac{2}{3}} d x \\
d y & =\frac{1}{3 \cdot x^{2 / 3}} d x=\frac{1}{3 .(27)^{\frac{2}{3}}}(-1) \\
\Delta y & =\frac{-1}{3(9)}=-\frac{1}{27} \approx-0.037 \\
3 \sqrt{26} & =y+\Delta y=3+(-0.037)=2.962
\end{aligned}
$
Hence, option (3) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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