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)=\frac1{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)+\frac x{3(1+0)^(2/3)}`

`L(x)=1+\frac x3`

which means we have


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


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

Compare to the actual values of


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


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