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
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http://prntscr.com/8bljw7

- tmagloire1

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- schrodinger

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- anonymous

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

- tmagloire1

a parabola

- tmagloire1

upward

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## More answers

- tmagloire1

@satellite73

- anonymous

a parabola that opens up, right
and the zeros are?

- tmagloire1

-1 and -2

- anonymous

ok good so lets draw a quick picture

- anonymous

|dw:1441157312300:dw|

- anonymous

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|

- tmagloire1

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

- anonymous

no it has a relative math right at \(x=-2\)
by a similar argument it has a relative min at \(x=-1\)

- anonymous

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

- tmagloire1

you lost me..

- anonymous

ok lets go slow then

- anonymous

you know what the derivative looks like right?

- tmagloire1

yes

- anonymous

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

- tmagloire1

yes i learned that

- anonymous

ok good
now "positive" is a synonym for "above the x axis"

- anonymous

|dw:1441157806983:dw|

- anonymous

that means on the interval
\[(-\infty,-2)\] and \[(-1,\infty)\]the function is increasing

- anonymous

not the derivative, the original function \(f\)
also on \((-2,-1)\) the function is decreasing

- anonymous

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

- tmagloire1

Ohh okay so you just us eincreasing/decreasing rules/ i understand thanks

- anonymous

yw

- tmagloire1

Can you help me with my second question as well?

- anonymous

sure but i didn't see it

- anonymous

oh now i do
take the derivative
what do you get?

- anonymous

i hope you get \[t^2-15t+50\]

- tmagloire1

For the derivative I got :t^2-50t+50

- tmagloire1

oops i meant 15

- anonymous

ok so your job is to solve \[t^2-15t+50<0\]

- anonymous

because if the derivative is negative, it is moving to the left

- anonymous

fortunately for you this factors, so it should be easy

- tmagloire1

t<5

- anonymous

is it clear how to solve that, or no?

- tmagloire1

t>10

- anonymous

i think you solved for where it is positive

- tmagloire1

How am I supposed to solve this then? The two intervals i got were those

- anonymous

it is zero at 5 and 10 right?

- anonymous

|dw:1441158937764:dw|

- tmagloire1

yes i got this

- anonymous

since this parabola opens up, it is negative between the zeros and positive outside of them
i.e negative on \((5,10)\)

- tmagloire1

is moving left another name for negative?

- anonymous

moving to the left = derivative is negative

- anonymous

i.e. the velocity is negative

- tmagloire1

so 5

- anonymous

yes

- tmagloire1

Thank you for explaining them to me! I appreciate it!

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