Question 889661
For an even function, {{{f(x)=f(-x)}}}
I. {{{3y=x^4+x}}}
{{{f(x)=(1/3)(x^4+x)}}}
{{{f(-x)=(1/3)(x^4-x)}}}
So I is not an even function.
{{{graph(300,300,-5,5,-5,5,(1/3)(x^4+x))}}}

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II. I assume you mean {{{f(x)=abs(x-2)*abs(x)}}}
{{{f(-x)=abs(-x-2)*abs(-x)}}}
Although {{{abs(x)=abs(-x)}}}, it's not the case that,
{{{abs(-x-2)=abs(x-2)}}}
So II is not an even function. 
It is symmetric but not about the y-axis but about {{{x=2}}}.
{{{graph(300,300,-5,5,-5,5,abs(x-2)*abs(x)(1/3))}}}
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III. {{{f(x)=e^(x^2)-1}}}
{{{f(-x)=e^(x^2)-1}}}
So III is an even function.
{{{graph(300,300,-5,5,-5,5,e^(x^2)-1)}}}
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The solution is c.