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Q.Find the approximate value of the cube root of 127.

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2020Subjective· 2mImportance★★★★★
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Use the linear (differential) approximation f(x+Δx)≈f(x)+f′(x) Δxf(x+\Delta x)\approx f(x)+f'(x)\,\Delta x with f(x)=x1/3f(x)=x^{1/3} at the nearby perfect cube x=125x=125.

Let f(x)=x1/3f(x)=x^{1/3}. Choose x=125x=125 (since 53=1255^3=125 is close to 127127) and Δx=127−125=2\Delta x=127-125=2.

f(125)=1251/3=5f(125)=125^{1/3}=5

f′(x)=13x−2/3⇒f′(125)=13(125)−2/3=13⋅125=175f'(x)=\dfrac13x^{-2/3}\quad\Rightarrow\quad f'(125)=\dfrac13(125)^{-2/3}=\dfrac13\cdot\dfrac{1}{25}=\dfrac{1}{75}

(using 1252/3=(1251/3)2=52=25125^{2/3}=\left(125^{1/3}\right)^2=5^2=25).

Linear approximation: …

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