Chapter 1.6, Problem 57E

To determine

To graph:The function $f\left(x\right)=\sqrt{x^3+x^2+x+1}$ and reason for the graph to be one to one, then find an explicit expression $f^{-1}\left(x\right)$ .

Explanation of Solution

Consider the function.

  $f\left(x\right)=\sqrt{x^3+x^2+x+1}$

The graph of the above function is shown in figure below.

  

Figure (1)

In the above graph, no horizontal line can be draw that intersects the graph more than once.

Thus, the function is one to one function.

To get the inverse replace the $x$ with $y$ and vice-versa in the given function.

  $x=\sqrt{y^3+y^2+y+1}$

With the help of CAS the solution obtained is,

  $y=\left[\begin{array}{l}-\frac{\left(1+i\sqrt{3}\right)\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}{6\sqrt[3]{2}}+\\ \frac{\sqrt[3]{2}\left(1-i\sqrt{3}\right)}{\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}-\frac{1}{3}\end{array}\right]$

Also,

  $y=\left[\begin{array}{l}-\frac{\left(1+i\sqrt{3}\right)\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}{6\sqrt[3]{2}}+\\ \frac{\sqrt[3]{2}\left(1-i\sqrt{3}\right)}{\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}-\frac{1}{3}\end{array}\right]$

And,

  $y=\left[\begin{array}{l}-\frac{\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}{3\sqrt[3]{2}}+\\ \frac{\sqrt[3]{2}}{\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}-\frac{1}{3}\end{array}\right]$

The two different solutions are not valid because they contain $i$ that makes them non-rea; solutions

Therefore, the inverse function is $f^{-1}\left(x\right)=\left[\begin{array}{l}-\frac{\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}{3\sqrt[3]{2}}+\\ \frac{\sqrt[3]{2}}{\sqrt[3]{27x^2+3\sqrt{3}\sqrt{27x^4-40x^2+16}-20}}-\frac{1}{3}\end{array}\right]$ .