Approximate value of \( \sqrt[3]{345} \), when it is calculated with the application of derivatives, is
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Always pick the closest known integer power point (\( 7^3 = 343 \)) as your reference base. Keeping your increment \( \Delta x \) small ensures the accuracy of your first-order differential approximation.
Step 1: Use the approximation rule. For a small change, $f(x+\Delta x)\approx f(x)+f'(x)\,\Delta x$. Choose a nearby perfect cube as the base. Step 2: Pick the base point. Let $f(x)=x^{1/3}$. The closest perfect cube to 345 is $343=7^3$, so $x=343$ and $f(343)=7$. Step 3: Find the increment. \[ \Delta x=345-343=2 \] Step 4: Differentiate. \[ f'(x)=\frac{1}{3}x^{-2/3}=\frac{1}{3(x^{1/3})^2} \] At $x=343$: $f'(343)=\frac{1}{3(7)^2}=\frac{1}{147}$. Step 5: Apply the formula. \[ f(345)\approx 7+\frac{1}{147}\times2=7+\frac{2}{147} \] Step 6: Compute the decimal. \[ \frac{2}{147}\approx0.0136\implies f(345)\approx7.013 \] \[ \boxed{7.013} \]