Question

Can you help me on this algebra 2 question?

Answer 1

Answer:

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

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

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

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

Step-by-step 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

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

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

$(f + g)( - 4) = - 10 \times - 2 = 20$

Then also;

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

$(f - g)( - 4) = 12 \times - 2 = - 24$

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.