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f:R→R, f(x)= [tex] x^{2009} [/tex] - 2009(x-1)-1
calculati f(0) + f'(0)

Răspuns :

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[tex]f(x)= x^{2009} - 2009(x-1)-1\Rightarrow f(0)=2008\\ f'(x)=2009x^{2008}-2009\Rightarrow f'(0)=-2009\\ f(0)+f'(0)=2008-2009=-1[/tex]

Obs.[tex](x^n)'=nx^{n-1}\Rightarrow (x^{2009})'=2009x^{2008}[/tex]

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