World of Functions

**magentarita** · January 5th 2009, 10:13 PM

Determine Algebraically if the functions are even, odd or neither.

(1) F(x) = cuberoot{x}...I said neither. Is this correct?

(2) F(x) = 2x/|x|...I said odd. Is this correct?

**Mush** · January 5th 2009, 10:38 PM

Originally Posted by magentarita

Determine Algebraically if the functions are even, odd or neither.

(1) F(x) = cuberoot{x}...I said neither. Is this correct?

(2) F(x) = 2x/|x|...I said odd. Is this correct?

You are correct about (2) but (1) is odd.

Because:

$f(x) = x^{\frac{1}{3}}$

$-f(-x) = -1(-x)^{\frac{1}{3}}$
$= -1((-1)^{\frac{1}{3}}(x)^{\frac{1}{3}})$
$= ((-1)^3(-1))^{\frac{1}{3}}x^{\frac{1}{3}}$
$=((-1)(-1))^{\frac{1}{3}}x^{\frac{1}{3}}$
$= 1^{\frac{1}{3}}x^{\frac{1}{3}}$
$=x^{\frac{1}{3}}$

By definition, a function is odd if $f(x) = -f(-x)$

**magentarita** · January 6th 2009, 06:14 AM

Originally Posted by Mush

You are correct about (2) but (1) is odd.

Because:

$f(x) = x^{\frac{1}{3}}$

$-f(-x) = -1(-x)^{\frac{1}{3}}$
$= -1((-1)^{\frac{1}{3}}(x)^{\frac{1}{3}})$
$= ((-1)^3(-1))^{\frac{1}{3}}x^{\frac{1}{3}}$
$=((-1)(-1))^{\frac{1}{3}}x^{\frac{1}{3}}$
$= 1^{\frac{1}{3}}x^{\frac{1}{3}}$
$=x^{\frac{1}{3}}$

By definition, a function is odd if $f(x) = -f(-x)$

I forgot to examine the cuberoot of {x} as x raised to the 1/3. Interesting reply. Thanks.

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