Approximate sqrt[3]{23.5} by means of differentials or a linear approximation. Use for the base point the closest integer that is convenient with regard to the cube root function. Warning: WeBWorK wants the approximation found using the method described above; a root read off a calculator display will be counted as wrong.

Question

anonymous · Answer

It looks like you're asked to find an approximation for $\sqrt[3]{23.5}$; if that's the case, I would suggest using $a=27$ as the closest integer.

So let $f(x)=\sqrt[3]x$. The linear approximation at a point $c=23.5$ is given by
$$f(c)\approx f'(c)(c-a)+f(a)$$
So,
$$f(23.5)\approx f'(23.5)(23.5-27)+f(27)$$
Of course, $f(x)=\sqrt[3]x~~\Rightarrow~~f'(x)=\dfrac{3}{4}x^{4/3}$.

anonymous · Answer

so do i solve for 3/4x^3/4 to get the right answer? or the equation above that ?

whpalmer4 · Answer

actually, that's not right.  $f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}$so $$f'(x) = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$$

whpalmer4 · Answer

So $$f'(23.5) = \frac{1}{3*(23.5)^{\frac{2}{3}}}$$

whpalmer4 · Answer

The result you should get if you evaluate all of that, when cubed, gives 23.3397, which is pretty close.  The actual value of $\sqrt[3]{23.5} \approx 2.86433$, for comparison purposes.  Note the warning about getting  a wrong answer if you are tempted to paste that in as your answer!