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Exercise 8.1 · Q1

Q.Let f(x)=x3f(x)=\sqrt[3]{x}. Find the linear approximation at x=27x=27. Use the linear approximation to approximate 27.23\sqrt[3]{27.2}.

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✓ Free question

We build the linear approximation L(x)=f(x0)+f′(x0)(x−x0)L(x)=f(x_0)+f'(x_0)(x-x_0) of f(x)=x3f(x)=\sqrt[3]x at the perfect cube x0=27x_0=27 (where f,f′f,f' are exact), then evaluate LL at x=27.2x=27.2.

Step 1. Set up ff and locate a convenient base point. f(x)=x1/3f(x)=x^{1/3}. Since 27=3327=3^3 is a perfect cube near 27.227.2, take x0=27x_0=27, so Δx=27.2−27=0.2\Delta x=27.2-27=0.2.

Step 2. Compute f(x0)f(x_0) and f′(x0)f'(x_0). f(27)=271/3=3f(27)=27^{1/3}=3. f′(x)=13x−2/3f'(x)=\dfrac13x^{-2/3}, so f′(27)=13⋅1272/3=13⋅19=127f'(27)=\dfrac13\cdot\dfrac{1}{27^{2/3}}=\dfrac13\cdot\dfrac19=\dfrac1{27} (using 272/3=(271/3)2=32=927^{2/3}=(27^{1/3})^2=3^2=9).

Step 3. Write the linear approximation. By Definition 8.1, L(x)=f(27)+f′(27)(x−27)=3+127(x−27)L(x)=f(27)+f'(27)(x-27)=3+\dfrac1{27}(x-27).

Step 4. Evaluate at x=27.2x=27.2. L(27.2)=3+127(0.2)=3+0.227=3+1135L(27.2)=3+\dfrac1{27}(0.2)=3+\dfrac{0.2}{27}=3+\dfrac1{135}.

Step 5. Simplify. 1135≈0.007407\dfrac1{135}\approx0.007407, so 27.23≈3.0074\sqrt[3]{27.2}\approx3.0074 (the true value is 27.23≈3.00739\sqrt[3]{27.2}\approx3.00739, an excellent match).

✓Final answer

L(x)=3+127(x−27)L(x)=3+\dfrac{1}{27}(x-27), and 27.23≈3+1135≈3.0074\sqrt[3]{27.2}\approx3+\dfrac{1}{135}\approx\boxed{3.0074}.

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