What is the relative maximum and minimum of the function? f(x) = 2x3 + x2 - 11
Having trouble with this. The relative is deferent then the local correct? Sorry to bother you again, if your still up, have any idea on this one? @satellite73 @Luigi0210
has a min, not max since it is a parabola that opens up min is the second coordinate of the vertex
oh crap it is cubic!! sorry
\[f(x) = 2x^3 + x^2 - 11\] \[f'(x)=3x^2+2x\] set it equal to zero and solve it has a relative max and a relative min
it is cubic so it goes from \(-\infty\) to \(\infty\) and has no absolute max or min
if you solve \[3x^2+2x=0\] you get \(x=-\frac{2}{3}\) or \(x=0\)
the one on the left will give a relative max and \(0\) will give a relative min
answers
oh and no relative is local same thing
oh damn i made a mistake
\[f'(x)=6x^2+2x\]
but none of your answer are answers to this one
I mean its from my high school I think there has to be one
check the function again, i am quite positive these are not answers to that one
f(x) = 2x^3 + x^2 - 11
did you know that?
\[f(x) = 2x^3 + x^2 - 11\] \[f'(x)=6x^2+2x\] for sure now set \(6x^2+2x=0\) you get \[2x(3x+1)=0\] so either \(x=0\) or \(x=-\frac{1}{3}\)
your answers are not to this question, or someone made a mistake sorry
hmm interesting. if you had to make a guess?
they are not even close don't guess, they are wrong
well its for online highschool and if i dont check any then its definitely wrong. Heres a pic of the whole question. Maybe you miss understood something...
yeah i see it it is wrong
relative min is \(-11\) at \(x=0\)
look i know this is on line and you have to put in something, but you should contact them somehow and tell them it is wrong don't give in to the machine anyway, if you have a real teacher he or she will think you are paying attention and be happy
the answers are not even close
okay i sent a message in. Thanks
I saw you answered a similar question. What polynomial has a graph that passes through the given points? (-2, 2) (-1, -1) (1, 5) (3, 67)
@satellite73
@Mathard
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