Question

Determine the linear approximation for f(x)=\sqrt[3]{x} at x=27 . Use the linear approximation to approximate the value of \sqrt[3]{28.45} . Please enter your answer in decimal format

Answer 1

The linear approximation for
`f(x)=\sqrt[3]{x}` at x=27 is approximately 3.4833, and this can be used to estimate
`\sqrt[3]{28.45}`.

The linear approximation for a function f(x) at a point x=a is given by the formula:

`L(x)=f(a)+f^(\prime)(a)(x-a)`

For the function
`f(x)=\sqrt[3]{x}` , we first need to find the derivative
`f^(\prime)(x)`. Let's compute it:

`\begin{aligned}&f(x)=x^(1 / 3)\\&f^(\prime)(x)=(1)/(3) x^(-2 / 3)\end{aligned}`

Now, we want to find the linear approximation at x=27

`\begin{aligned}& L(x)=f(27)+f^(\prime)(27)(x-27) \\& L(x)=3+(1)/(3)(x-27)\end{aligned}`

Now, to approximate
`\sqrt[3]{28.45}` , we use x=28.45 in the linear approximation:

`\begin{aligned}& L(28.45)=3+(1)/(3)(28.45-27) \\& L(28.45)=3+(1)/(3)(1.45) \\& L(28.45)=3+0.4833 \\& L(28.45) \approx 3.4833\end{aligned}`

So, the linear approximation for
`\sqrt[3]{28.45}` is approximately 3.4833 in decimal format.

Answer 2

The linear approximation is

`3 + (x)/(27) - 1`

The value of

`\sqrt[3]{28.45}`

is 3.054.

How to determine approximated value of a function

Given

`f(x) = \sqrt[3]{x}`

To linearize the function at x = 27

L(x) = f(a) + f'(a) * (x - a)

L(x) is the linear approximation function.

f(x) is the original function.

a is the point around which we are linearizing.

f'(a) is the derivative of f(x) evaluated at a.

x is the variable.

`f(x) = {x}^{ (1)/(3) }`

`= \sqrt[3]{27 } = 3`

Find f'(x)

`(dy)/(dx) = (1)/(3) {x}^{ (1)/(3) - 1}`

`(dy)/(dx) = (1)/(3) {x}^{ - (2 )/(3) }`

`= (1)/(3) {27}^{ - (2)/(3) } = (1)/(3) { (\sqrt[3]{27} }^( ) - 2)`

`= (1)/(27)`

L(x) = f(a) + f'(a) * (x - a)

Substitute

`= 3 + (1)/(27) * (x - 27)`

`= 3 + (x)/(27) - 1`

To approximate

`\sqrt[3]{28.45}`

x = 28.45

Therefore

`l(28.45) = 3 + (28.45)/(27) - 1`

= 3 + 1.0536 -1

= 3.054

Therefore, approximated value is 3.054