Question

question 3 of 15 (4 points) | question attempt: 1 of unlimited
(a) given ( f(x) = 6x^2 + 2|x| ), find ( f(-x) ).
( f(-x) = 6(-x)^2 + 2|(-x)| )
( = 6x^2 + 2|x| )
part 2 of 3
(b) is ( f(-x) = f(x) )?
yes, ( f(-x) = f(x) )
part: 2 / 3
part 3 of 3
(c) is this function even, odd, or neither?
since ( f(-x) = f(x) ), the function is select

Explanation:

Step1: Substitute $x$ with $-x$

$f(-x)=6(-x)^2 + 2|-x|$

Step2: Simplify squared and absolute value

$(-x)^2=x^2$, $|-x|=|x|$, so $f(-x)=6x^2 + 2|x|$

Step3: Compare $f(-x)$ and $f(x)$

$f(x)=6x^2 + 2|x|$, so $f(-x)=f(x)$

Step4: Classify the function

By definition, if $f(-x)=f(x)$, the function is even.

Answer:

(a) $6x^2 + 2|x|$
(b) Yes, $f(-x)=f(x)$
(c) Even