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f(x) = -3 e^x + 2 + e^-5 find f'(0)
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\[\Large\bf\sf f(x)\quad=\quad-3e^{x}+2+e^{-5}\]Have you tried finding f'(x) yet?
Did I type that correctly? Or is the +2 supposed to be in the exponent also?
-3 e^(x+2) is f'(x)
yeah +2 supposed to be in the exponent too
\[f(x)=-3e ^{x+2}+e ^{-5}=-3e ^{2}e ^{x}+e ^{-5}\] \[f'(x)=-3e ^{2}e ^{x},put ~x=0~and~get~the~result.\]
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thanks, i got it :D
Whenever you come across function notation that tells you to find f(4) (or something), always plug that into X. F(x)=1x^2+12 F(4)=1(4)^2+12 :)
got chu
bless you.
That was a sneeze right? 0_o
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LOL too funny
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