Chapter 2.3, Problem 38E

To determine

To find: The reason for the continuity of the function $y=x^2+\sqrt[3]{4-x}$ , check the domain of the function, also mention the theorems used in the continuity of function with the functions assumed as continuous.

Answer to Problem 38E

The domain of the function $y=x^2+\sqrt[3]{4-x}$ is equal to interval $\left(-\infty ,\infty \right)$ and the sum, composition theorems are used and the functions $x^2$ , $\sqrt[3]{x}$ and $4-x$ are assumed as continuous.

Explanation of Solution

Given information:

The function is $y=x^2+\sqrt[3]{4-x}$ .

Calculation:

The function $y=x^2+\sqrt[3]{4-x}$ is the sum of two functions $f\left(x\right)=x^2$ and $g\left(x\right)=\sqrt[3]{4-x}$ .

The function $y=x^2+\sqrt[3]{4-x}$ is equal to the function $f\left(x\right)+g\left(x\right)$ .

Assume the function $f\left(x\right)=x^2$ is a continuous function.

The function $g\left(x\right)=\sqrt[3]{4-x}$ is the composition of two functions $h\left(x\right)=\sqrt[3]{x}$ and $j\left(x\right)=4-x$ .

Assume that the function $h\left(x\right)=\sqrt[3]{x}$ and $j\left(x\right)=4-x$ are continuous functions.

So, by the composition theorem the function $g\left(x\right)=\left(hoj\right)\left(x\right)$ is continuous function.

So, by the help of sum theorem the function $y=x^2+\sqrt[3]{4-x}$ is continuous as the functions $f\left(x\right)$ and $g\left(x\right)$ both are continuous.

The function $y=x^2+\sqrt[3]{4-x}$ is defined for all the real values of $x$ . So, the domain of the function $y=x^2+\sqrt[3]{4-x}$ is equal to interval $\left(-\infty ,\infty \right)$ .

Therefore, the domain of the function $y=x^2+\sqrt[3]{4-x}$ is equal to interval $\left(-\infty ,\infty \right)$ and the sum, composition theorems are used and the functions $x^2$ , $\sqrt[3]{x}$ and $4-x$ are assumed as continuous.