Find values of a and b such that the function

f(x)=ax+b /(x^2)−1
has a local extreme value of 1 at x=3. Is this extreme value a local maximum, or a local minimum? Give reasons for your answer.

Question

Find values of a and b such that the function

f(x)=ax+b /(x^2)−1
has a local extreme value of 1 at x=3. Is this extreme value a local maximum, or a local minimum? Give reasons for your answer.

freckles · Answer

so first let's try to find f'

anonymous · Answer

will do.

freckles · Answer

also is it (ax+b)/(x^2-1)?

anonymous · Answer

yup

zale101 · Answer

Find the derivative and set it equal to zero. After solving for x, those will be the critical points. Do you know how to take the derivative of this function?

anonymous · Answer

$$\frac{ ax+b }{ x^2 -1 }$$ = f(x)

anonymous · Answer

that's the original, and to get the derivative i use quotient rule.

anonymous · Answer

So i think it's :
$$\frac{ -ax^2 -a -2bx }{ (x^2 -1)^2 }$$

freckles · Answer

you will have a system of equations to solve by the way
$$f(3)=1 \ \text{ then set the numerator of } f' =0 \text{ then solve for x}\  \text{ then use that that x is suppose to be 3}$$

freckles · Answer

himm I think I got a little bit of a different numerator

freckles · Answer

oh you combined like terms lol

freckles · Answer

sounds good

freckles · Answer

$$-ax^2-2bx-a=0$$
solve for x

freckles · Answer

use the quadratic formula

anonymous · Answer

okay let's see...

anonymous · Answer

I got:
$$x = a + or - {\sqrt{a^2-8abx}}$$

anonymous · Answer

it's supposed to be above -2a

anonymous · Answer

$$\frac{ a  + or - \sqrt {a^2 - 8abx}  }{ -2a}$$

freckles · Answer

$$-ax^2-2bx-a=0 \ x=\frac{-(-2b) \pm \sqrt{(-2b)^2-4(-a)(-a)}}{2(-a)} \ x=\frac{2b \pm \sqrt{4b^2-4a^2}}{-2a}\ x=\frac{b \pm \sqrt{b^2-a^2}}{-a} \ \text{ we want one of these x's to be 3 }$$
so we have
$$\frac{b \pm \sqrt{b^2-a^2}}{-a}=3 $$
and we also have the other equation:
from f(3)=1
$$f(3)=\frac{3a+b}{9-1}=\frac{3a+b}{8}=1 \ 3a+b=8  \ b=8-3a$$
so i would plug this b in the ther other equation

freckles · Answer

and hopefully we can solve for a

anonymous · Answer

oh i think i was using quadtratic on the wrong thing no clue where the x came from

freckles · Answer

and don't worry the a's do not look that bad
i got integers

anonymous · Answer

so i plug b into the quadtratic formula answer?

freckles · Answer

yeah replace the b's with (8-3a)

freckles · Answer

having trouble?

anonymous · Answer

it was a small mistake during my calculations again. sorry

anonymous · Answer

|dw:1418439005597:dw|
so like this?