Find values of a and b so that the function f(x) = (ax+b)/(x^2-1) has a local extreme value of 1 at x=3. is this value a local minimum or local maximum?

Answer 1

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I hope this helps!

Answer 2

Hi Aisha T.,

Sholdn't be hard to differentiate that function, use the chain rule, etc. Then set f ' = 0, and solve.

I make it: -(ax+b)*(x^2-1)^-2 * 2x + (x^2-1)^-1 * a = f '

(-2ax^2-2bx)*(x^2-1)^-2 + a*(x^2-1)^-1 = 0 multiply through by (x^2-1)^2:

(-2ax^2-2bx) = -a(x^2-1)

2(ax^2-bx) = ax^2 - a

ax^2 - 2bx + a = 0

but, we don't need to solve for x, x=3! So 9a -6b +a = 0 ; 10a = 6b so b = (5/3)a

Plug back in to f(x) to satisfy the boundary condition:

f(3)=1:

(3a+(5/3)a)/(9-1) = 1 (14/3)a/8 = 1 a= 24/14 = 12/7 b= (12/7)*(5/3) = 20/7

to find out if a maximum or a minimum could either figure f ' ' or calculate a few points, since (x^2-1) isn't varying much in this interval ....

IFF I haven't made any math bloopers??

--Cheers, --Mr. d.