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Q.Using differentials, find the approximate value of 26^(1/3) upto 3 places of decimal. OR Find the interval in which the function f(x) = x² + 2x - 5 is strictly increasing or decreasing.

Himachal HpboseHPBOSE Plus Two Board 2022Subjective· 3mImportance★★★★★
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Use y=x1/3y=x^{1/3} near x=27x=27 (a perfect cube close to 26) with the differential dy≈dydx dxdy\approx \dfrac{dy}{dx}\,dx.

Let y=x1/3y=x^{1/3}. Take x=27x=27 (since 271/3=327^{1/3}=3 is exact and close to 26) and Δx=dx=26−27=−1\Delta x=dx=26-27=-1.

dydx=13x−2/3\dfrac{dy}{dx}=\dfrac13 x^{-2/3}

At x=27x=27: dydx=13(27)−2/3=13⋅19=127\dfrac{dy}{dx}=\dfrac13(27)^{-2/3}=\dfrac13\cdot\dfrac{1}{9}=\dfrac{1}{27}

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