Rolle's Theorem question? :) [Calculus]?

Question

Determine if Rolle's Theorem can be applied to
f(x)=(x²+3x-18)/(x+9) on the interval (-6,3), and if it can, find all numbers c satisfying the conclusion of that theorem.

When I worked it out, I got c=-3,-15 and Rolle's Theorem not applicable. Both turned out to be wrong :( Thanks for the help!

Answer 1

Rolle's Theorem is applicable since f(-6) = 0 and f(3) = 0.
So using the theorm there must be a number c such that f'(c) = 0.

f'(x) = [(2x+3)*(x+9) - (x²+3x-18)* 1]/(x+9)² = 0
(2x²+21x+27 - x²-3x+18)/(x+9)² = 0
-x²-18x+45 / (x+9)² = 0

Answer 2

Firstly, Rolle's Theorem is applicable since f(-6) = f(3) and is continuous on (-6,3). Note that there is a discontinuity at x= -9, but this isn't in the interval and doesn't matter.

Secondly, you found the critical points of -3 and -15, but -15 is not in the interval (-6,3) and is not a consequence of Rolle's Theorem, and hence is not a value "satisfying the conclusion of that theorem."

Answer 3

f(-6)= 0
(f(3) = 0
The function is contiunuos except for x=-9 outside the interval and the derivative exists
so Rolle ism appicable and
f´(x) =0 at least at some point interior
f´(x)= 1/(x+9)^2 *[(x+9)(2x+3)-(x^2+3x-18)]
x^2+18x+45=0
x=((-18+-sqrt(144)/2 so x= -3 and x=-15 outside the intervall
so the only point is x=-3

Answer 4

f(x)=(x²+3x-18)/(x+9)
f(-6)=((-6)²+3(-6)-18)/((-6)+9)
f(-6)=(36-18-18)/3
f(-6)=0/3
f(-6)=0/3

f(3)=(3²+3*3-18)/(3+9)
f(3)=(9+9-18)/(3+9)
f(3)=0/12
f(3)=0

So, f(-6)=f(3) – the first requirement of Rolle's Theorem is met.

Next, f must be differentiable on the interior of the interval, i,e., (-6, 3).

f'(x)=(x²+3x-18)/(x+9)
f'(x)=([d(x²+3x-18)/dx]*(x+9)-(x²+3x-18)*[d(x+9)/dx])/(x+9)²
f'(x)=[(2x+3)(x+9)-(x²+3x-18)*1]/(x+9)²
f'(x)=(2x²+21x+27-x²-3x+18)/(x+9)²
f'(x)=(x²+18x+45)/(x+9)²
Thus, f is differentiable on (-6, 3).

By Rolle's Theorem, there must be at least one value a in (-6, 3) such that f(a)=0. In particular,

0=(x²+18x+45)/(x+9)²
0=x²+18x+45
x=(-18+/-√18²-4*1*45)/(2*1)
x=(-18+/-√324-180)/2
x=(-18+/-√144)/2
x=(-18+/-12)/2
x=-30/2 or -6/2
x=-15 or -3

Of these roots, only -3 lies on the interior of (-6, 3).