tmagloire1
  • tmagloire1
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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
do you know what \(y=(x+1)(6+3x)\) looks like?
tmagloire1
  • tmagloire1
a parabola
tmagloire1
  • tmagloire1
upward

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

tmagloire1
  • tmagloire1
@satellite73
anonymous
  • anonymous
a parabola that opens up, right and the zeros are?
tmagloire1
  • tmagloire1
-1 and -2
anonymous
  • anonymous
ok good so lets draw a quick picture
anonymous
  • anonymous
|dw:1441157312300:dw|
anonymous
  • 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
  • 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
  • anonymous
no it has a relative math right at \(x=-2\) by a similar argument it has a relative min at \(x=-1\)
anonymous
  • 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
  • tmagloire1
you lost me..
anonymous
  • anonymous
ok lets go slow then
anonymous
  • anonymous
you know what the derivative looks like right?
tmagloire1
  • tmagloire1
yes
anonymous
  • 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
  • tmagloire1
yes i learned that
anonymous
  • anonymous
ok good now "positive" is a synonym for "above the x axis"
anonymous
  • anonymous
|dw:1441157806983:dw|
anonymous
  • anonymous
that means on the interval \[(-\infty,-2)\] and \[(-1,\infty)\]the function is increasing
anonymous
  • anonymous
not the derivative, the original function \(f\) also on \((-2,-1)\) the function is decreasing
anonymous
  • 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
  • tmagloire1
Ohh okay so you just us eincreasing/decreasing rules/ i understand thanks
anonymous
  • anonymous
yw
tmagloire1
  • tmagloire1
Can you help me with my second question as well?
anonymous
  • anonymous
sure but i didn't see it
anonymous
  • anonymous
oh now i do take the derivative what do you get?
anonymous
  • anonymous
i hope you get \[t^2-15t+50\]
tmagloire1
  • tmagloire1
For the derivative I got :t^2-50t+50
tmagloire1
  • tmagloire1
oops i meant 15
anonymous
  • anonymous
ok so your job is to solve \[t^2-15t+50<0\]
anonymous
  • anonymous
because if the derivative is negative, it is moving to the left
anonymous
  • anonymous
fortunately for you this factors, so it should be easy
tmagloire1
  • tmagloire1
t<5
anonymous
  • anonymous
is it clear how to solve that, or no?
tmagloire1
  • tmagloire1
t>10
anonymous
  • anonymous
i think you solved for where it is positive
tmagloire1
  • tmagloire1
How am I supposed to solve this then? The two intervals i got were those
anonymous
  • anonymous
it is zero at 5 and 10 right?
anonymous
  • anonymous
|dw:1441158937764:dw|
tmagloire1
  • tmagloire1
yes i got this
anonymous
  • anonymous
since this parabola opens up, it is negative between the zeros and positive outside of them i.e negative on \((5,10)\)
tmagloire1
  • tmagloire1
is moving left another name for negative?
anonymous
  • anonymous
moving to the left = derivative is negative
anonymous
  • anonymous
i.e. the velocity is negative
tmagloire1
  • tmagloire1
so 5
anonymous
  • anonymous
yes
tmagloire1
  • tmagloire1
Thank you for explaining them to me! I appreciate it!

Looking for something else?

Not the answer you are looking for? Search for more explanations.