How do you use a linear approximation to approximate the value of $\sqrt[3]{27.5}$ $root(3)(27.5)$ ?

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Answer 1 · 2017-08-23T17:47:12

We let $f (x) = \sqrt[3]{a}$ $f(x) = root(3)(a)$ . Then $f (27) = \sqrt[3]{27} = 3$ $f(27) = root(3)(27) = 3$

We compute the derivative of $f (x)$ $f(x)$ as being

$f(a) = a^(1/3) -> f'(a) = 1/3a^(-2/3)$

Hence,

$f'(27) = 1/3(27)^(-2/3) =1/3(3)^(-2) = 1/3(1/9) = 1/27$

Recalll the equation of a line is given by

$y - y_1 = m(x - a)$

$y - 3 = 1/27(x - 27)$

$y = 1/27x - 1+ 3$

$y = 1/27x + 2$

So the approximation at $x = 27.5$ would be

$y ~~ 1/27(27.5) + 2 ~~ 3.0185185...$

If we use a calculator to compute we get $3.0184054$ apprxoaimtely, so this approximation is very good.

Hopefully this helps!

How do you use a linear approximation to approximate the value of root(3)(27.5)3√27.5root(3)(27.5)?