Chapter 2.1, Problem 74E

Solution

To analyse the function $f\left(x\right)=x^2$ for its domain, range, continuity, increasing-decreasing behaviour, symmetry, boundedness, local extrema, horizontal asymptotes, vertical asymptotes and end behaviour.

The domain of the function $f\left(x\right)=x^2$ :

  $Domain\left(f\left(x\right)\right)=ℝ$ .

The range of the function $f\left(x\right)=x^2$ :

  $Range\left(f\left(x\right)\right)=\left[0,\infty \right)$ .

The function $f\left(x\right)=x^2$ is continuous everywhere.

The function $f\left(x\right)=x^2$ is strictly increasing when $x>0$ and it is strictly decreasing when $x<0$ .

The function $f\left(x\right)=x^2$ has only the lower bound. Also, one of its bounds is $0$ .

The function $f\left(x\right)=x^2$ has only a local minimum and it is attained at $x=0$ .

The function $f\left(x\right)=x^2$ has no any horizontal asymptotes.

The function $f\left(x\right)=x^2$ has no any vertical asymptotes.

The function $f\left(x\right)=x^2$ approaches positive infinity as $x$ approaches infinity on either side.

Given the function:

  $f\left(x\right)=x^2$

Concept used:

Given a function $f\left(x\right)$ .

Domain of the function:

The domain of the function: $Domain\left(f\left(x\right)\right)=\left\{x\in ℝ|f\left(x\right)\text{is}\text{defined}\right\}$ .

The range of the function:

Also, the range of the function: $Range\left(f\left(x\right)\right)=\left\{f\left(x\right)\in ℝ|x\in Domain\left(f\left(x\right)\right)\right\}$ .

Theorem on differentiability and continuity:

If a function is differentiable then it is continuous.

Increasing and decreasing behaviour of the function provided its derivative:

The function $f\left(x\right)$ will be strictly increasing if $f^{\text{'}}\left(x\right)>0$ and it will be strictly decreasing if $f^{\text{'}}\left(x\right)<0$ . Also, it will be a constant when $f^{\text{'}}\left(x\right)=0$ .

Symmetry of an even function:

An even function will always be symmetric about the $y$ -axis.

Boundedness of a function:

The function $f\left(x\right)$ is said to be bounded if there exists $m,M\in ℝ$ so that $m\le f\left(x\right)\le M$ .

In this case, $m$ is said to be the lower bound of the function and $M$ is said to be the upper bound of the function.

Local extrema of a function:

The function $f\left(x\right)$ will have its local extrema at its critical points at which either $f^{\text{'}}\left(x\right)=0$ or $f^{\text{'}}\left(x\right)$ is undefined.

If $f^{\text{'}}\left(a\right)=0$ and if $f^{\text{'}\text{'}}\left(a\right)>0$ , then $x=a$ is a local minimum of $f\left(x\right)$ .

If $f^{\text{'}}\left(a\right)=0$ and if $f^{\text{'}\text{'}}\left(a\right)<0$ , then $x=a$ is a local maximum of $f\left(x\right)$ .

Horizontal asymptote of a function:

The horizontal asymptotes are horizontal lines that the graph of the function approaches as $x\to \pm \infty$ .

That is, the horizontal asymptotes are:

  $y=\underset{x\to -\infty }{\lim }f\left(x\right)$ and $y=\underset{x\to \infty }{\lim }f\left(x\right)$ .

Vertical asymptotes of a function:

The line $x=a$ is a vertical asymptote of the function $f\left(x\right)$ if $\underset{x\to a^{-}}{\lim }f\left(x\right)=\pm \infty$ or $\underset{x\to a^{+}}{\lim }f\left(x\right)=\pm \infty$ .

End Behaviour of a function:

The end behaviour of the function $f\left(x\right)$ is what it does when $x$ approaches infinity on either side.

Analysis:

The domain of the function $f\left(x\right)=x^2$ :

  $\begin{array}{l}Domain\left(f\left(x\right)\right)=\left\{x\in ℝ|f\left(x\right)\text{is}\text{defined}\right\}\\ Domain\left(f\left(x\right)\right)=\left\{x\in ℝ|x^2\text{is}\text{defined}\right\}\\ Domain\left(f\left(x\right)\right)=ℝ\end{array}$ .

Thus, the domain of $f\left(x\right)=x^2$ is $ℝ$ .

The range of the function $f\left(x\right)=x^2$ :

  $\begin{array}{l}Range\left(f\left(x\right)\right)=\left\{f\left(x\right)\in ℝ|x\in Domain\left(f\left(x\right)\right)\right\}\\ Range\left(f\left(x\right)\right)=\left\{x^2\in ℝ|x\in ℝ\right\}\\ Range\left(f\left(x\right)\right)=\left[0,\infty \right)\end{array}$ .

Thus, the range of $f\left(x\right)=x^2$ is $\left[0,\infty \right)$ .

Analysing the continuity of the function $f\left(x\right)=x^2$ :

Observe that $f\left(x\right)=x^2$ is differentiable everywhere with the derivative $f^{\text{'}}\left(x\right)=2x$ .

Hence, it is continuous everywhere.

Analysing the increasing and decreasing behaviour of the function $f\left(x\right)=x^2$ :

Observe that $f^{\text{'}}\left(x\right)=2x$ is positive when $x>0$ and that $f^{\text{'}}\left(x\right)=2x$ is negative when $x<0$ .

Hence, the function $f\left(x\right)=x^2$ is strictly increasing when $x>0$ and it is strictly decreasing when $x<0$ .

Analyse the symmetry of the function:

Observe that the function $f\left(x\right)=x^2$ is an even function. Hence, it is symmetric about the $y$ -axis.

Analyse and find the bounds of the function:

For $x\in ℝ$ , $x^2\in \left[0,\infty \right)$ .

That is:

  $0\le f\left(x\right)\le \infty$ .

Thus, the function has only lower bound and the number $0$ is such a lower bound.

Find the local extrema of the function:

It is already said that the function $f\left(x\right)=x^2$ is differentiable everywhere with derivative $f^{\text{'}}\left(x\right)=2x$ .

So, there are no any points of non-differentiability.

Now, $f^{\text{'}}\left(x\right)=0\Rightarrow x=0$ .

That is, the function has only one critical point: $x=0$ .

Now, $f^{\text{'}\text{'}}\left(x\right)=2$ .

That is, $f^{\text{'}\text{'}}\left(0\right)=2>0$ .

Thus, $x=0$ is a point corresponding to a local minimum of $f\left(x\right)=x^2$ .

Find the horizontal asymptotes of the function:

Observe that:

  $\begin{array}{c}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }{x}^2\\ \underset{x\to -\infty }{\lim }f\left(x\right)=\infty \end{array}$ .

Also, observe that:

  $\begin{array}{c}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{x}^2\\ \underset{x\to \infty }{\lim }f\left(x\right)=\infty \end{array}$ .

That is, the function diverges to positive infinity as $x$ approaches infinity on either side.

That is, the function has no any horizontal asymptote.

Find the vertical asymptotes of the function:

Observe that the function $f\left(x\right)=x^2$ has no any $x=a$ so that $\underset{x\to a^{-}}{\lim }f\left(x\right)=\pm \infty$ or $\underset{x\to a^{+}}{\lim }f\left(x\right)=\pm \infty$ .

That is, the function has no any vertical asymptotes.

Analyse the end behaviour of the function:

It may be observed that the function approaches infinity as $x$ approaches infinity on either side.

Conclusion:

The domain of the function $f\left(x\right)=x^2$ :

  $Domain\left(f\left(x\right)\right)=ℝ$ .

The range of the function $f\left(x\right)=x^2$ :

  $Range\left(f\left(x\right)\right)=\left[0,\infty \right)$ .

The function $f\left(x\right)=x^2$ is continuous everywhere.

The function $f\left(x\right)=x^2$ is strictly increasing when $x>0$ and it is strictly decreasing when $x<0$ .

The function $f\left(x\right)=x^2$ has only the lower bound. Also, one of its bounds is $0$ .

The function $f\left(x\right)=x^2$ has only a local minimum and it is attained at $x=0$ .

The function $f\left(x\right)=x^2$ has no any horizontal asymptotes.

The function $f\left(x\right)=x^2$ has no any vertical asymptotes.

The function $f\left(x\right)=x^2$ approaches positive infinity as $x$ approaches infinity on either side.