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How do I find the domain of

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(4x^3-.5x+2)/(2x^3+.3x^2-1)? Is the function odd, even, or either?

asked Nov 20, 2014 in PRECALCULUS by anonymous

2 Answers

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The function f(x) = (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

We know that all possible values of x is domain of a function.

A rational function is simply fraction and in a fraction the denominator cannot be equal to 0 because it would be undefined.

To find which number make the fraction undefined create an equation where the denominator is not equal to zero.

2x3 + 3x2 - 1 ≠ 0

2x3 + 2x2 + x2 - 1 ≠ 0

2x2(x + 1) + (x + 1)(x - 1) ≠ 0

(x + 1)[ 2x2 + (x - 1)] ≠ 0

(x + 1)(2x2 + x - 1) ≠ 0

(x + 1)(2x2+ 2x - x - 1) ≠ 0

(x + 1)[2x(x + 1) - 1(x + 1)] ≠ 0

(x + 1)[(x + 1) (2x - 1)] ≠ 0

(x + 1)(x + 1)(2x - 1)] ≠ 0

x + 1 ≠ 0 and 2x - 1 ≠ 0

x ≠ - 1 and x ≠ 1/2

So the domain of the function all real numbers except - 1 and 1/2.

Domain set is {x∈R: x ≠ -1 and x ≠ 1/2}.

 

f(x) = (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

Find the f(- x) and - f(x).

 f(- x) = [4(- x)3 - 5(- x) + 2]/[2(- x)3 + 3(- x)2 - 1]

= ( - 4x3 + 5x + 2)/(- 2x3 + 3x2 - 1)

= - (4x3 - 5x - 2)/-(2x3 - 3x2 + 1)

=  (4x3 - 5x - 2)/(2x3 - 3x2 + 1)

f(x) ≠ f(- x) so f is not even.

- f(x) = - (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

= (- 4x3 + 5x - 2)/(- 2x3 - 3x2 + 1)

= - (4x3 - 5x + 2)/-(2x3 + 3x2 - 1)

= (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

f(- x) ≠ - f(x) , so f is not odd.

f(x) = (4x3 - 5x + 2)/(2x3 + 3x2 - 1) is neither even nor odd.

answered Nov 21, 2014 by david Expert
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f(x) = (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

Find the f(- x) and - f(x).

 f(- x) = [4(- x)3 - 5(- x) + 2]/[2(- x)3 + 3(- x)2 - 1]

= ( - 4x3 + 5x + 2)/(- 2x3 + 3x2 - 1)

= - (4x3 - 5x - 2)/-(2x3 - 3x2 + 1)

=  (4x3 - 5x - 2)/(2x3 - 3x2 + 1)

f(x) ≠ f(- x) so f is not even.

- f(x) = - (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

= (- 4x3 + 5x - 2)/(- 2x3 - 3x2 + 1)

= - (4x3 - 5x + 2)/-(2x3 + 3x2 - 1)

= (4x3 - 5x + 2)/(2x3 + 3x2 - 1)

f(- x) ≠ - f(x) , so f is not odd.

f(x) = (4x3 - 5x + 2)/(2x3 + 3x2 - 1) is neither even nor odd.

answered Nov 21, 2014 by david Expert

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