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 2.1, Problem 18E
To determine

Evaluate the given function f(x)=x3+2x , by substitute the different numbers for x .

Expert Solution

Answer to Problem 18E

The valuesof the given function are :

  f(2)=12,f(1)=3,f(0)=0,f(13)=1927,f(0.2)=0.408

Explanation of Solution

Given information: consider the function f(x)=x3+2x

Calculation:

Evaluate the values of the function f(x)=x3+2x for the given values of the variable x

Plug the value x=2 in the function f(x)=x3+2x then we get the value of f(2).

  f(2)=(2)3+2(2)f(2)=84f(2)=12 Substitute x=2

Add 8 and 4

Plug the value x=1, in the function f(x)=x3+2x then we get the value of f(1).

   f( 1 )= ( 1 ) 3 +2( 1 ) f( 1 )=1+2 f( 1 )=3

Substitute x=1

Add 1 and 2

Plug the value x=0, in the function f(x)=x3+2x then we get the value of f(0).

  f(0)=(0)3+2(0)f(0)=0+0f(0)=0

Substitute x=0

Add 0 and 0

Plug the value x=13 in the function f(x)=x3+2x then we get the value of f(13).

   f( 1 3 )= ( 1 3 ) 3 +2( 1 3 ) f( 1 3 )= 1 27 + 2 3 f( 1 3 )= 1+18 27 f( 1 3 )= 19 27

Substitute x=13

Add 127 and 23

Find least mean square

Plug the value x=0.2 in the function f(x)=x3+2x then we get the value of f(0.2).

  f(0.2)=(0.2)3+2(0.2)f(0.2)=0.008+0.4f(0.2)=0.408

   Substitute x=0.2

Add 0.08 and 0.4

Thus, the values of the given function are: f(2)=12,f(1)=3,f(0)=0,f(13)=1927,f(0.2)=0.408

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