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Find the relative extrema: f(x) = x^4+4x^3
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\[f'(x)=4x^3+12x^2\] \[4x^2(x+3)\] \[x=0,-3\]
yeah thats what i got:\
but at the back the answer is only: Rel. Min(-3,-27)
hmmm
have you try to plug in -3 into original equation/
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would that be a mistake my teacher made
if you plug in -3 into original equation you should get -27
I got -189 for some reason...
@winterfez
\[(-3)^4+4(-3)^3\]
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use BRACKET for -3
OHHH
IN'T THAT REL. MAX THOUGH? not rel. min
oops sorry for CAPS xD
is (-3,-27) the Relative Max not Relative Min?
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and the Relative min. would be 0 rite?
It not really specify if they want an interval or not
-27 is relative min at poin -3 and 0 is max at point 0
oh wait nvm
thank you :)
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or you could say -27 is relative min at x=-3...0 is max at x=0
YW
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