Question: H: g[h (x)] = E: glg (x)] = G: h[g(x)] = D: g[f (x)] = C: f [f (x)] = 1: 'h [h (x) ]
![H: g[h (x)] = E: glg (x)] = G: h[g(x)] = D: g[f](https://dsd5zvtm8ll6.cloudfront.net/si.experts.images/questions/2025/02/67a7e28c6e0c2_75667a7e28c5139e.jpg)
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