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 1.9, Problem 36E
To determine

To calculate: The value of x from the equation x3+16=0 by algebraic method and graphical method.

Expert Solution

Answer to Problem 36E

The value of x is (16)13=2.519 and Graph of the equation is provided below,

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 36E , additional homework tip  1

Explanation of Solution

Given information:

The equation x3+16=0 .

Calculation:

It is provided that the equation x3+16=0 .

Algebraic method;

The method in which isolate x on one side of the equation and solve the value of x .

From the equation x3+16=0 .

Subtract 16 on the both side of the equation,

  x3+1616=16

Solve the value of x from the equation,

  x3=16x=1613x=2.519

Therefore, the value of x=2.519 .

Graphical method;

Let put y1=x3 and y2=16

The graph of the equation y1=x3 provided below,

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 36E , additional homework tip  2

The graph of equation y2=16 is provided below,

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 36E , additional homework tip  3

The graph of both equations in same figure is provided below,

  Precalculus: Mathematics for Calculus - 6th Edition, Chapter 1.9, Problem 36E , additional homework tip  4

The solution is the x coordinate of the intersection point of the two graphs. So, the intersection point on the x coordinate is 2.519 .

Thus, the value of x is 2.519 .

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