Chapter 2.1, Problem 73E

Solution

The complete analysis of given function.

Domain: $\left(-\infty ,\infty \right),\left\{x\mid x\in ℝ\right\}$

Range: $\left(-\infty ,\infty \right),\left\{y\mid y\in ℝ\right\}$

Continuity: $f\left(x\right)=x$ is continuous over all real numbers.

Increasing-decreasing behaviour: The graph is always increasing.

Symmetry: The function is odd, so, the graph is symmetric about the origin.

Boundedness: Upper and lower bound $=0$

Local extrema: No Local Extrema

Horizontal asymptotes: No Asymptotes

Vertical asymptotes: No Asymptotes

End behaviour: Th graph falls to the left and rises to the right

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

Graph:

  

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

Set-Builder Notation: $\left\{x|x\in ℝ\right\}$

Range:

The range is the set of all valid $y$ values. Use the graph to find the range.

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

Set-Builder Notation: $\left\{y\mid y\in ℝ\right\}$

Continuity:

Since the domain is all real numbers, $f\left(x\right)=x$ is continuous over all real numbers.

Increasing-decreasing behaviour:

The function $f\left(x\right)=x$ is always increasing as shown in the figure.

Symmetry:

To determine if the function is odd, even, or neither in order to find the symmetry.

If the function is odd, the function is symmetric about the origin. And ff the function is even, the function is symmetric about the y-axis.

Find $f\left(-x\right)$ by substituting $-x$ for all occurrence of $x$ in $f\left(x\right)$

  $f\left(-x\right)=-x$

A function is even if $f\left(-x\right)=f\left(x\right)$

Check if $f\left(-x\right)=f\left(x\right)$

  $-x\ne x$

Since $-x\ne x$ , the function is not even.

The function is not even.

A function is odd if $f\left(-x\right)=-f\left(x\right)$

Multiply $-1$ by $x$ :

  $-f\left(x\right)=-x$

Since $-x=-x$ , the function is odd.

The function is odd.

Since the function is odd, the graph is symmetric about the origin.

Boundedness:

Find every combination of $\pm \frac{p}{q}.$

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

  $\begin{array}{l}p=\pm 0\\ q=\pm 1\end{array}$

Find every combination of $\pm \frac{p}{q}$ . Thus, the possible roots of the polynomial function is $0.$

Apply synthetic division on $\frac{x}{x}$ when $x=0$

  $\begin{array}{l}0\overline{)\begin{array}{l}10\\ 0\end{array}}\\ 10\end{array}$

Since $0>0$ and all of the signs in the bottom row of the synthetic division are positive, $0$ is the upper and lower bound for the real roots of the function.

Upper and lower bound: $0$

Local extrema:

Differentiate $f\left(x\right)=x$ using the Power Rule which states that $\frac{d}{dx}\left[x^n\right]$ is $n{x}^{n-1}$ where $n=1$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

  $f^{\prime \prime }\left(x\right)=0$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ :

  $1=0$

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

Horizontal asymptotes:

Since $y=x$ is not a rational expression, there are no asymptotes.

Vertical asymptotes:

Since $y=x$ is not a rational expression, there are no asymptotes.

End behaviour:

The largest exponent is the degree of the polynomial is $1.$ Thus, the leading coefficient is $1.$

Since the degree is odd, the ends of the function will point in the opposite directions. Falls to the left and rises to the right

Since the leading coefficient is positive, the graph rises to the right.

From analysing the graph, the function falls to the left and rises to the right