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F(x)=x-1 supra e^x
F(0)+f'(0) =?


Răspuns :

Salut,

[tex]f(0)=0-\dfrac{1}{e^0}=-\dfrac{1}1=-1;\\\\f'(x)=\left(x-\dfrac{1}{e^x}\right)'=(x-e^{-x})'=1-(-1)\cdot e^{-x}=e^{-x}+1;\\\\f'(0)=e^{0}+1=1+1=2;\\f(0)+f'(0)=-1+2=1.[/tex]

Simplu, nu ?

Green eyes.