Question

Can you help me on this algebra 2 question?

Answer 1

`(f + g)(x) = - 10 \sqrt[3]{2x} \: \: \: (f + g)( - 4) = 20`

`(f - g)(x) =12 \sqrt[3]{2x} \: \: \: (f - g)( - 4) = - 24`

The domain of (f+g)(x) is all real numbers.

The domain of (f-g)(x) is all real numbers.

Explanation:

The given functions are

`f(x) = \sqrt[3]{2x}`

and

`g(x) =- 11\sqrt[3]{2x}`

By the algebraic properties of polynomial functions:

`(f + g)(x) = f(x) + g(x)`

`(f + g)(x) = \sqrt[3]{2x} + - 11 \sqrt[3]{2x}`

This becomes:

`(f + g)(x) = \sqrt[3]{2x} - 11 \sqrt[3]{2x}`

We subtract to obtain:

`(f + g)(x) = - 10 \sqrt[3]{2x}`

Also

`(f - g)(x) = f(x) - g(x)`

`(f - g)(x) = \sqrt[3]{2x} - - 11 \sqrt[3]{2x}`

`(f - g)(x) = \sqrt[3]{2x} + 11 \sqrt[3]{2x}`

`(f - g)(x) = 12\sqrt[3]{2x}`

When x=-4

`(f + g)( - 4) = - 10\sqrt[3]{2 * - 4}`

`(f + g)( - 4) = - 10\sqrt[3]{ - 8}`

`(f + g)( - 4) = - 10 * - 2 = 20`

Then also;

`(f - g)( - 4) = 12\sqrt[3]{ - 8}`

`(f - g)( - 4) = 12 * - 2 = - 24`

The domain refers to the values that makes the function defined.

Both are cube root functions and are defined for all real numbers.

The domain of (f+g)(x) is all real numbers.

The domain of (f-g)(x) is all real numbers.