Question

question
fill out the table of values and select whether each function is odd, even or neither.
$f(x) = 7x^3 + 5x^5$

$x$	$y$
$-1$
$0$
$1$
$2$

odd
$f(x) = 2|x| - 1$

$x$	$y$
$-1$
$0$
$1$
$2$

even
$f(x) = 7x^5 + 5$

$x$	$y$
$-1$
$0$
$1$
$2$

neither
answer attempt 1 out of 2
you must answer all questions above in order to submit.

Explanation:

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For $f(x)=7x^3 + 5x^5$

Step1: Calculate $f(-2)$

$f(-2)=7(-2)^3 + 5(-2)^5 = 7(-8)+5(-32)=-56-160=-176$

Step2: Calculate $f(-1)$

$f(-1)=7(-1)^3 + 5(-1)^5 = 7(-1)+5(-1)=-7-5=-12$

Step3: Calculate $f(0)$

$f(0)=7(0)^3 + 5(0)^5 = 0+0=0$

Step4: Calculate $f(1)$

$f(1)=7(1)^3 + 5(1)^5 = 7+5=12$

Step5: Calculate $f(2)$

$f(2)=7(2)^3 + 5(2)^5 = 7(8)+5(32)=56+160=176$

Step6: Verify odd function

Check $f(-x)=-f(x)$: $f(-2)=-f(2)$, $f(-1)=-f(1)$, $f(0)=0$, so it is odd.

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For $f(x)=2|x| - 1$

Step1: Calculate $f(-2)$

$f(-2)=2|-2| -1=2(2)-1=4-1=3$

Step2: Calculate $f(-1)$

$f(-1)=2|-1| -1=2(1)-1=2-1=1$

Step3: Calculate $f(0)$

$f(0)=2|0| -1=0-1=-1$

Step4: Calculate $f(1)$

$f(1)=2|1| -1=2(1)-1=2-1=1$

Step5: Calculate $f(2)$

$f(2)=2|2| -1=2(2)-1=4-1=3$

Step6: Verify even function

Check $f(-x)=f(x)$: $f(-2)=f(2)$, $f(-1)=f(1)$, so it is even.

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For $f(x)=7x^5 + 5$

Step1: Calculate $f(-2)$

$f(-2)=7(-2)^5 +5=7(-32)+5=-224+5=-219$

Step2: Calculate $f(-1)$

$f(-1)=7(-1)^5 +5=7(-1)+5=-7+5=-2$

Step3: Calculate $f(0)$

$f(0)=7(0)^5 +5=0+5=5$

Step4: Calculate $f(1)$

$f(1)=7(1)^5 +5=7+5=12$

Step5: Calculate $f(2)$

$f(2)=7(2)^5 +5=7(32)+5=224+5=229$

Step6: Verify classification

$f(-2)
eq -f(2)$ and $f(-2)
eq f(2)$, so it is neither.

Answer:

For $f(x)=7x^3 + 5x^5$:

$x$	$y$
$-1$	$-12$
$0$	$0$
$1$	$12$
$2$	$176$

Classification: Odd

For $f(x)=2|x| - 1$:

$x$	$y$
$-1$	$1$
$0$	$-1$
$1$	$1$
$2$	$3$

Classification: Even

For $f(x)=7x^5 + 5$:

$x$	$y$
$-1$	$-2$
$0$	$5$
$1$	$12$
$2$	$229$

Classification: Neither