Question

Find a linear approximation of the function f(x)=\sqrt[3]{1+x} at a=0, and use it to approximate the numbers \sqrt[3]{.96} and \sqrt[3]{1.02}

Answer 1

$f(x)=\sqrt[3]{1+x}=(1+x)^{1/3}\implies f'(x)=\dfrac1{3(1+x)^{2/3}}$

The linear approximation to $f(x)$ around $x=a$ is then

$L(x)=f(a)+f'(a)(x-a)\approx f(x)$

So the approximation centered at $a=0$ will be

$L(x)=f(0)+f'(0)x=(1+0)^{1/3}+\dfrac x{3(1+0)^{2/3}}$
$L(x)=1+\dfrac x3$

which means we have

$\sqrt[3]{0.96}\approx L(0.96)=1+\dfrac{0.96}3=1.32$

$\sqrt[3]{1.02}\approx L(1.02)=1+\dfrac{1.02}3=1.34$

Compare to the actual values of

$\sqrt[3]{0.96}\approx0.9864$

$\sqrt[3]{1.02}\approx1.0066$