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Question Number 12359 by sin (x) last updated on 20/Apr/17

Answered by ajfour last updated on 20/Apr/17

$$forgivenrangeof\mathit{a}and\mathit{b},$$$${({\mathit{a}}^4+{\mathit{b}}^3)}_{\mathit{m}\mathit{a}\mathit{x}}=1-\frac{1}{8}=\frac{7}{8}$$$${({\mathit{a}}^4+{\mathit{b}}^3)}_{\mathit{m}\mathit{i}\mathit{n}}=0+(-8+h)=-8+h$$$$\mathit{s}\mathit{o}\left(\mathit{c}\right)\to (-8,\frac{7}{8}].$$

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