Christos
  • Christos
Graphs and functions/derivatives, Hey can you help me with this? http://screencast.com/t/MVj3e9Ok2Yg0
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
.Sam.
  • .Sam.
\[p(x)=(x+1)^2(2x-x^2)\] From there, you can tell that (x+1)=0, x=-1 is a zero of p(x) Then factor this \((2x-x^2)\) and set the brackets equals zero and find the other 2 zeros
Christos
  • Christos
its -1 0 and 2
Christos
  • Christos
Oh so thats the first step

Looking for something else?

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

More answers

.Sam.
  • .Sam.
Yup |dw:1369655367345:dw|
.Sam.
  • .Sam.
Now, differentiate p(x), set it equals to zero to find the maximum and minimum points
Christos
  • Christos
its -4x^3+6x+2 if I am not mistaken we take x=0 and x=+-sqrt(3)/sqrt(2)
.Sam.
  • .Sam.
Yeah, set -4x^3+6x+2 equals zero and solve for x
Christos
  • Christos
yea I solved for x and I got 3 roots x=0 and x=+-sqrt(3)/sqrt(2) are my calculations correct??
.Sam.
  • .Sam.
Hmm, \[-4x^3+6x+2 \\ \\ -2(2x^3-3x-1) \\ \\ -2(x+1)(2x^2-2x-1)\] Factoring \(2x^2-2x-1\) you get \[x=\frac{1}{2} \left(1-\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\] These x are the max and min points of the graph, then use these x substitute into \[y=(x+1)^2(2x-x^2)\] Find the y's
Christos
  • Christos
hmm how did you get from line 2 to line 3?
Christos
  • Christos
http://screencast.com/t/yNU8TQxrMkn this here
.Sam.
  • .Sam.
I used trial an error with calculator, when \[p(-1)=-2(2(-1)^3-3(-1)-1)=0\] Then (x+1) is a factor of p(x), then use (x+1) divide \((2x^3-3x-1)\)
1 Attachment
Christos
  • Christos
oh ok lets find the ys
Christos
  • Christos
y = 0 ?
Christos
  • Christos
I subtitute with 0 or the roots?
.Sam.
  • .Sam.
Substitute these \(x=\frac{1}{2} \left(1-\sqrt{3}\right)~~~~~,~~~~~\frac{1}{2} \left(1+\sqrt{3}\right)\) into \(y=(x+1)^2(2x-x^2)\)
.Sam.
  • .Sam.
Then that's the coordinates of max and minimum points
Christos
  • Christos
that was really long to find wasnt it? I tried to find the first one and I found (-2 + sqrt(3))/2 which seems to be wrong here
.Sam.
  • .Sam.
Yeah It's abit long For \(x=\frac{1}{2} \left(1-\sqrt{3}\right)\) or -0.3660, \[y=(x+1)^2(2x-x^2)\] \[y=(\frac{1-\sqrt3}{2} +1)^2(2(\frac{1-\sqrt3}{2})-(\frac{1-\sqrt3}{2})^2)\] \[y=\frac{9-6\sqrt{3}}{4}~~or-0.348\]
Christos
  • Christos
ok I see
.Sam.
  • .Sam.
(-0.3660,-0.3481) Then for the other coordinate is (1.366,4.848)
Christos
  • Christos
ok
Christos
  • Christos
Now that we know those coordinates how can we use them?
.Sam.
  • .Sam.
Yup |dw:1369657703814:dw|
.Sam.
  • .Sam.
Then the coefficient of highest degree is negative so the graph curves down|dw:1369657774047:dw|
.Sam.
  • .Sam.
And yeah, zeros are -1 0 and 2, why am I plotting 1 for.
Christos
  • Christos
hmm are you plotting it to get an estimation of the infliction point?
.Sam.
  • .Sam.
Yup
.Sam.
  • .Sam.
or else you won't know where it changes it's direction
Christos
  • Christos
I see , is there anything else left for this graph/exersise?
.Sam.
  • .Sam.
That's it
Christos
  • Christos
Thanks :) I will try to solve some on my own now, if I have any problem can I tell you?
.Sam.
  • .Sam.
I'll try :)
Christos
  • Christos
ok

Looking for something else?

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