BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 1.6, Problem 57E
To determine

To graph:The function f(x)=x3+x2+x+1 and reason for the graph to be one to one, then find an explicit expression f1(x) .

Expert Solution

Explanation of Solution

Consider the function.

  f(x)=x3+x2+x+1

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

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 1.6, Problem 57E

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=y3+y2+y+1

With the help of CAS the solution obtained is,

  y=[(1+i3)27x2+3327x440x2+16203623+23(1i3)27x2+3327x440x2+1620313]

Also,

  y=[(1+i3)27x2+3327x440x2+16203623+23(1i3)27x2+3327x440x2+1620313]

And,

  y=[27x2+3327x440x2+16203323+2327x2+3327x440x2+1620313]

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

Therefore, the inverse function is f1(x)=[27x2+3327x440x2+16203323+2327x2+3327x440x2+1620313] .

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