Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 29

$$a = 512,{\text{ }}\frac{1}{{\root 3 \of {510} }} \approx 0.1251627$$

$$\eqalign{ & \frac{1}{{\root 3 \of {510} }} \cr & {\text{Rewrite }}\frac{1}{{\root 3 \of {510} }}{\text{ as }}\frac{1}{{\root 3 \of {512 - 2} }} \cr & {\text{Using the function }}f\left( x \right) = \frac{1}{{\root 3 \of x }},{\text{ }} \cr & f\left( x \right) = \frac{1}{{\root 3 \of x }} \cr & {\text{we have that }}510 = 512 - 2,{\text{ let }}a = 512 \cr & {\text{* Evaluate }}f\left( {512} \right) \cr & f\left( {512} \right) = \frac{1}{{\root 3 \of {512} }} \cr & f\left( {512} \right) = \frac{1}{8} \cr & \cr & {\text{Differentiating we obtain}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\root 3 \of x }}} \right] = \frac{d}{{dx}}\left[ {{x^{ - 1/3}}} \right] \cr & f'\left( x \right) = - \frac{1}{3}{x^{ - 4/3}} \cr & {\text{* Evaluate }}f'\left( {512} \right) \cr & f'\left( {512} \right) = - \frac{1}{3}{\left( {512} \right)^{ - 4/3}} \cr & f'\left( {512} \right) = - \frac{1}{{12288}} \cr & \cr & {\text{Using the definition of a linear approximation to }}f{\text{ at }}a{\text{ }} \cr & L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right),{\text{ for an interval }}I{\text{ containing }}a \cr & \cr & {\text{The linear approximation }}L\left( x \right){\text{ at }}a{\text{ is:}} \cr & L\left( x \right) = \frac{1}{8} - \frac{1}{{12288}}\left( {x - 512} \right) \cr & \cr & {\text{Estimating the given value at }}f\left( {510} \right) \cr & L\left( {510} \right) = \frac{1}{8} - \frac{1}{{12288}}\left( {510 - 512} \right) \cr & L\left( {510} \right) = 0.1251627 \cr & {\text{Therefore}}{\text{,}} \cr & \frac{1}{{\root 3 \of {510} }} \approx L\left( {510} \right) \approx 0.1251627 \cr} $$