BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 3.1, Problem 56E
To determine

To find: The local maximum and minimum values of the function f(x)=3+x+x2x3 and also the values of x on which it occurs.

Expert Solution

Answer to Problem 56E

The local maximum value of the function f(x)=3+x+x2x3 is equal to 4 at x=1 and the local minimum value of the function f(x)=3+x+x2x3 is equal to 2.815 at x=0.333 .

Explanation of Solution

Given information:

The given function is f(x)=3+x+x2x3 .

Calculation:

Graph the given function f(x)=3+x+x2x3 .

To graph a function y=3+x+x2x3 , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press Y= key and enter right hand side of the equation y=3+x+x2x3 after the symbol Y1 . Enter the keystrokes 3+X+X^2X^3 .

The display will show the equation,

  Y1=3+X+X^2X^3

Now, press the GRAPH key and TRACE key to produce the graph of given function in standard window as shown in Figure (1).

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 3.1, Problem 56E

Figure (1)

As observed from the graph, the function has local maximum value equal to 4 at x=1 .

As observed from the graph, the function has local minima value equal to 2.815 at x=0.333 .

Therefore, the local maximum value of the function f(x)=3+x+x2x3 is equal to 4 at x=1 and the local minimum value of the function f(x)=3+x+x2x3 is equal to 2.815 at x=0.333 .

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