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I think it isn't because for x = 2 and x = 4, h(x) is not defined. Is that right?

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yowhatsup123
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    What does "valid" mean to you here? You are correct that it is undefined for $x\in {2,4}$. – lulu Mar 17 '22 at 18:31
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    Agreed, yes. This would be fine if the domain were $\Bbb{R} \setminus {2, 4}$ instead, but since the domain is very clearly marked as $\Bbb{R}$, you would expect a well-defined value at $x = 2$ and $x = 4$. – Theo Bendit Mar 17 '22 at 18:47

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It is not a valid function, because despite the domain being indicated to be $\mathbb{R},$ there is no element $y$ in the codomain such that $(2,y)\in{h}$ and the same is true for $(4,y).$

Angel
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