geometrical applications of calculus (1 Viewer)

blacktown_ · Nov 16, 2012

evaluate a and b if y=x^3-ax^2+bx-3 has stationary points at x=-1 and x=2

please help!!!!!

thankyou

Carrotsticks · Nov 16, 2012

superSAIyan2 · Nov 16, 2012

Differentiate with respect to x
dy/dx = 3x^2 - 2ax + b
since x=-1 and 2 are stat pts y'=0
0 = 3 + 2a + b (1)
0 = 12 - 4a + b (2)

Then just solve simultaneously.
(1) - (2) 6a = 9 a = 3/2
when a =3/2 b = -6 (dunno if this is right, did it in my head )

Capt Rifle · Nov 16, 2012

damn beat me to it haha

blacktown_ · Nov 16, 2012

Thankyou everyone that answered

Currybear · Nov 16, 2012

How did you get your text like that, im kind of new..

Carrotsticks said:
$\\ y=x^3-ax^2+bx-3 \\\\ y' = 3x^2 - 2ax + b = 0 $ for stationary points.$ \\\\ $Since it has stationary points at $ x=-1 $ and $ x=2 $, we can substitute in those values into $ y' $ and it will be equal to zero.$ \\\\ $Substitute $ x=-1: \\\\ 3+2a+b=0 \qquad (1) \\\\ $Substitute $ x=2: \\\\ 12-4a+b=0 \qquad (2) \\\\ $Now we can solve these two expressions simultaneously.$ \\\\ (2)-(1): \\\\ 9-6a=0 \Rightarrow a=\frac{3}{2} \\\\ $Substitute into (1):$ \\\\ 3 + 2 \times \frac{3}{2} + b = 0 \\\\ 3 + 3 + b = 0 \Rightarrow b=-6$

Carrotsticks · Nov 16, 2012

Currybear said:
How did you get your text like that, im kind of new..

It's called LaTeX and takes a bit of time to learn to use properly.

http://community.boredofstudies.org/showthread.php?t=234262

lizzy95 · Jan 9, 2013

thank you this helped so much!

geometrical applications of calculus (1 Viewer)

blacktown_

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