Question:
By application of derivatives, approximate value of
\[
\sqrt[5]{242}
\]
is
By application of derivatives, approximate value of
\[
\sqrt[5]{242}
\]
is
Show Hint
For approximation by derivatives, always choose the nearest number whose exact value is easy to compute.
Updated On: Jun 15, 2026
- \(2.9085\)
- \(2.9975\)
- \(2.9527\)
- \(2.8529\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
Approximation formula:
\[
f(a+h)\approx f(a)+hf'(a)
\]
Take function
\[
y=x^{1/5}
\]
Choose nearby perfect power.
Step 1: Choose nearest value.
\[ 243=3^5 \] Thus \[ 242=243-1 \] Take \[ f(x)=x^{1/5} \]
Step 2: Find derivative.
\[ f'(x)=\frac15x^{-4/5} \] At \[ x=243 \] \[ f'(243)=\frac1{5(3^4)} = \frac1{405} \]
Step 3: Apply approximation.
\[ f(242) \approx f(243)-f'(243) \] \[ = 3-\frac1{405} \] \[ = 3-0.002469 \] \[ = 2.99753 \] Hence \[ \boxed{2.9975} \]
Step 1: Choose nearest value.
\[ 243=3^5 \] Thus \[ 242=243-1 \] Take \[ f(x)=x^{1/5} \]
Step 2: Find derivative.
\[ f'(x)=\frac15x^{-4/5} \] At \[ x=243 \] \[ f'(243)=\frac1{5(3^4)} = \frac1{405} \]
Step 3: Apply approximation.
\[ f(242) \approx f(243)-f'(243) \] \[ = 3-\frac1{405} \] \[ = 3-0.002469 \] \[ = 2.99753 \] Hence \[ \boxed{2.9975} \]
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