Chapter 1.3, Problem 38E

To determine

To Verify: The given function is continuous and state the domain. State the theorem

used and the function that assumed to be continuous.

Answer to Problem 38E

The given function is continuous and it is verified by using the sum theorem,

composite theorem and product theorems.

The domain is $\left(-\infty ,\infty \right)$

Explanation of Solution

Given:

  $y=x^2+\sqrt[3]{4-x}$

Calculation:

Given the function $y=x^2+\sqrt[3]{4-x}$ :

Here $f\left(x\right)=x$ , $g\left(x\right)=4$ and $h\left(x\right)=\sqrt[3]{x}$ are continuous by using various theorems:

By using the sum theorem: $k\left(x\right)=g\left(x\right)-f\left(x\right)$

Substitute $f\left(x\right)=x$ and $g\left(x\right)=4$

  $\begin{array}{c}k\left(x\right)=f\left(x\right)+g\left(x\right)\\ =4-x\end{array}$ which is continuous.

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

By using the composite theorem, $\left(h\circ k\right)\left(x\right)=h\left(k\left(x\right)\right)$

Substitute $k\left(x\right)=4-x$ :

  $\begin{array}{c}\left(h\circ k\right)\left(x\right)=h\left(k\left(x\right)\right)\\ =\sqrt[3]{4-x}\end{array}$ which is continuous

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

And by using the product theorem: $f\left(x\right)\cdot f\left(x\right)$

Substitute $f\left(x\right)=x$ :

  $\begin{array}{c}f\left(x\right)\cdot f\left(x\right)=x\cdot x\\ =x^2\end{array}$ which is continuous.

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

And by using the sum theorem: $y=f{\left(x\right)}^2+\left(h\circ k\right)\left(x\right)$

Substitute $f\left(x\right)\cdot f\left(x\right)=x^2$ and $\left(h°k\right)\left(x\right)=\sqrt[3]{4-x}$

  $\begin{array}{c}y=f{\left(x\right)}^2+\left(h\circ k\right)\left(x\right)\\ =x^2+\sqrt[3]{4-x}\end{array}$ which is continuous.

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

Find Domain:

The domain of the expression is all real numbers except where the expression is

undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation: $\left(-\infty ,\infty \right)$

Thus, the domain of the function is $\left(-\infty ,\infty \right)$ .