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Exercise 8.1 · Q2

Q.Use the linear approximation to find approximate values of

(i) (123)2/3(123)^{2/3}
(ii) 154\sqrt[4]{15}
(iii) 263\sqrt[3]{26}
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For each part, choose a base point x0x_0 that is an exact perfect power close to the target, so f(x0)f(x_0) and f′(x0)f'(x_0) are both exact, then apply f(x0+Δx)≈f(x0)+f′(x0)Δxf(x_0+\Delta x)\approx f(x_0)+f'(x_0)\Delta x.

Part (i): (123)2/3(123)^{2/3}. Take f(x)=x2/3f(x)=x^{2/3}, x0=125=53x_0=125=5^3 (nearest perfect cube), Δx=123−125=−2\Delta x=123-125=-2.

f(125)=1252/3=(53)2/3=52=25f(125)=125^{2/3}=(5^3)^{2/3}=5^2=25. f′(x)=23x−1/3f'(x)=\dfrac23x^{-1/3}, so f′(125)=23⋅15=215f'(125)=\dfrac23\cdot\dfrac15=\dfrac2{15}.

(123)2/3≈25+215(−2)=25−415=375−415=37115≈24.733(123)^{2/3}\approx25+\dfrac2{15}(-2)=25-\dfrac4{15}=\dfrac{375-4}{15}=\dfrac{371}{15}\approx24.733.

Part (ii): 154\sqrt[4]{15}. Take f(x)=x1/4f(x)=x^{1/4}, x0=16=24x_0=16=2^4, Δx=15−16=−1\Delta x=15-16=-1.

f(16)=161/4=2f(16)=16^{1/4}=2. f′(x)=14x−3/4f'(x)=\dfrac14x^{-3/4}, so f′(16)=14⋅1163/4=14⋅18=132f'(16)=\dfrac14\cdot\dfrac{1}{16^{3/4}}=\dfrac14\cdot\dfrac18=\dfrac1{32} (since 163/4=(161/4)3=23=816^{3/4}=(16^{1/4})^3=2^3=8).

154≈2+132(−1)=2−132=64−132=6332≈1.96875\sqrt[4]{15}\approx2+\dfrac1{32}(-1)=2-\dfrac1{32}=\dfrac{64-1}{32}=\dfrac{63}{32}\approx1.96875.

Part (iii): 263\sqrt[3]{26}. Take f(x)=x1/3f(x)=x^{1/3}, x0=27x_0=27, Δx=26−27=−1\Delta x=26-27=-1.

From Question 1, f(27)=3, f′(27)=127f(27)=3,\,f'(27)=\dfrac1{27}.

263≈3+127(−1)=3−127=81−127=8027≈2.96296\sqrt[3]{26}\approx3+\dfrac1{27}(-1)=3-\dfrac1{27}=\dfrac{81-1}{27}=\dfrac{80}{27}\approx2.96296.

✓Final answer

(i) (123)2/3≈37115≈24.733(123)^{2/3}\approx\dfrac{371}{15}\approx\boxed{24.733} (ii) 154≈6332≈1.969\sqrt[4]{15}\approx\dfrac{63}{32}\approx\boxed{1.969} (iii) 263≈8027≈2.963\sqrt[3]{26}\approx\dfrac{80}{27}\approx\boxed{2.963}

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