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

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

Mathematics
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do you know what \(y=(x+1)(6+3x)\) looks like?
a parabola
upward

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a parabola that opens up, right and the zeros are?
-1 and -2
ok good so lets draw a quick picture
|dw:1441157312300:dw|
at \(x=-2\) 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|
exactly 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
no it has a relative math right at \(x=-2\) by a similar argument it has a relative min at \(x=-1\)
at any rate the derivative has degree two, so the original function has degree 3 with positive leading coefficient|dw:1441157542119:dw|, so you have a good idea that it looks like this
you lost me..
ok lets go slow then
you know what the derivative looks like right?
yes
and 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
yes i learned that
ok good now "positive" is a synonym for "above the x axis"
|dw:1441157806983:dw|
that means on the interval \[(-\infty,-2)\] and \[(-1,\infty)\]the function is increasing
not the derivative, the original function \(f\) also on \((-2,-1)\) the function is decreasing
it 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
Ohh okay so you just us eincreasing/decreasing rules/ i understand thanks
yw
Can you help me with my second question as well?
sure but i didn't see it
oh now i do take the derivative what do you get?
i hope you get \[t^2-15t+50\]
For the derivative I got :t^2-50t+50
oops i meant 15
ok so your job is to solve \[t^2-15t+50<0\]
because if the derivative is negative, it is moving to the left
fortunately for you this factors, so it should be easy
t<5
is it clear how to solve that, or no?
t>10
i think you solved for where it is positive
How am I supposed to solve this then? The two intervals i got were those
it is zero at 5 and 10 right?
|dw:1441158937764:dw|
yes i got this
since this parabola opens up, it is negative between the zeros and positive outside of them i.e negative on \((5,10)\)
is moving left another name for negative?
moving to the left = derivative is negative
i.e. the velocity is negative
so 5
yes
Thank you for explaining them to me! I appreciate it!

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