CALCULUS: STATIONARY POINT \[f(x)=x^3+Ax+B\] has a stationary point at (-2, 3). Find the values of A & B. --- I've already found A=-12. I just don't know how to find B.

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CALCULUS: STATIONARY POINT \[f(x)=x^3+Ax+B\] has a stationary point at (-2, 3). Find the values of A & B. --- I've already found A=-12. I just don't know how to find B.

Mathematics
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what does it mean if the function has a stationary point?
Stationary point means the derivative of that function is zero.
Well, I just know that if a function has a stationary point then the derivative of that function is equal to 0.

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oh... you already said that lol
Did you find b already or not?
Nope, not yet.
I don't know what to do.
Should I just plug in (2,3) to the function to solve for B?
If so, then I'll get B=-13. But I'm not sure.
this is not just about plugging in values you need to know what you're doing and why you're doing it
Yeah, that's why I am asking how to do it. :)
You should start with taking the derivative, set it equal to zero
how should not be the point of interest
Okay, I'll show my work.
set x=-2 and f(x) =3
^ that's what I did to find B.
|dw:1442978733930:dw|
Hmm..I don't think you can just leave out B
we know that in a graph |dw:1442978750479:dw|
a and b are constants...
@Jhannybean Well, B is a constant. SO the derivative is 0.
Oh, didn't see the derivative sign in your function, m'bad.
You didn't? That's fine. :D
After you solve for a and got -12 plug that to the original equation and dont forget to plug (-2,3) for x and f(x). From there, you can solve for b.
^ I did that and I got -13.
just like what I said a while ago :D
Okay, then you are good to go :)
Awesome! Thanks for the help though! :)
Anytime!
I actually just found out that stationary points are critical points based on this problem. I heart openstudy

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