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Question Number 212428 by Spillover last updated on 13/Oct/24

Answered by Yassine84 last updated on 13/Oct/24

$$letf\left(x\right)=(1+x^2)(1+x^3)(1+x^4)then$$$$li\underset{1}{m}\frac{(1+x^2)(1+x^3)(1+x^4)-8}{x-1}=f^{\prime }\left(1\right)=36$$$$$$$$because{f}^{\prime }\left(x\right)=2x(1+x^3)(1+x^4)+3x^2(1+x^2)(1+x^4)+4x^3(1+x^2)(1+x^3)$$$$$$

Answered by golsendro last updated on 13/Oct/24

$$=\underset{x\to 1}{\lim }\frac{(\mathrm{x}-1)({\mathrm{x}}^8+{\mathrm{x}}^7+{2\mathrm{x}}^6+{3\mathrm{x}}^5+{4\mathrm{x}}^4+{5\mathrm{x}}^3+{6\mathrm{x}}^2+7\mathrm{x}+7)}{\mathrm{x}-1}$$$$=1+1+2+3+4+5+6+7+7$$$$=8+\frac{7}{2}\left(8\right)=8+28=36$$

Answered by Nadirhashim last updated on 13/Oct/24

$$\mathit{e}\mathit{x}\mathit{p}\mathit{a}\mathit{n}\mathit{i}\mathit{g}\mathit{a}\mathit{n}\mathit{d}\mathit{d}\mathit{e}\mathit{v}\mathit{i}\mathit{d}$$$$\mathit{i}\mathit{m}({\mathit{x}}^8+x^7-2x^6+3x^5+4x^4+5x^3$$$$+4x^4+5x^3+6x^2+7x+7$$$$whenx=1theanswer=36$$

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