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tmagloire1
 one year ago
Given f '(x) = (x + 1)(6 + 3x), find the xcoordinate for the relative minimum on the graph of f(x).
0
−1
−2
None of these

http://prntscr.com/8bljw7
tmagloire1
 one year ago
Given f '(x) = (x + 1)(6 + 3x), find the xcoordinate for the relative minimum on the graph of f(x). 0 −1 −2 None of these  http://prntscr.com/8bljw7

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know what \(y=(x+1)(6+3x)\) looks like?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a parabola that opens up, right and the zeros are?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok good so lets draw a quick picture

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441157312300:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0at \(x=2\) the derivative goes from being positive (above the x axis) to negative (below the x axis) that tells you the function goes from increasing to decreasing at \(2\) i.e. up and then down therefore \(2\) is a relative max dw:1441157412113:dw

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0exactly what i drew. i got 1.5ish as a minimum but it wasn't an option so i wasn't sure if they were trying to trick me by putting 1 and 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no it has a relative math right at \(x=2\) by a similar argument it has a relative min at \(x=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0at any rate the derivative has degree two, so the original function has degree 3 with positive leading coefficientdw:1441157542119:dw, so you have a good idea that it looks like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok lets go slow then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know what the derivative looks like right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you know (or are supposed to be learning) that if the derivative is positive, then the function is increasing, and if the derivative is negative, tehn the function is decreasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok good now "positive" is a synonym for "above the x axis"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441157806983:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that means on the interval \[(\infty,2)\] and \[(1,\infty)\]the function is increasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not the derivative, the original function \(f\) also on \((2,1)\) the function is decreasing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it goes up until it gets to 2, then it goes down that means 2 gives a relative MAX it goes down until 1, then it goes back up again that means 1 gives a relative MIN

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Ohh okay so you just us eincreasing/decreasing rules/ i understand thanks

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Can you help me with my second question as well?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sure but i didn't see it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh now i do take the derivative what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i hope you get \[t^215t+50\]

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0For the derivative I got :t^250t+50

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so your job is to solve \[t^215t+50<0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because if the derivative is negative, it is moving to the left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0fortunately for you this factors, so it should be easy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it clear how to solve that, or no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think you solved for where it is positive

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0How am I supposed to solve this then? The two intervals i got were those

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is zero at 5 and 10 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441158937764:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since this parabola opens up, it is negative between the zeros and positive outside of them i.e negative on \((5,10)\)

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0is moving left another name for negative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0moving to the left = derivative is negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i.e. the velocity is negative

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0so 5<t<10 would be correct?

tmagloire1
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for explaining them to me! I appreciate it!
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