Question

Determine whether Rolle's Theorem can be applied to f on the indicated interval. If Rolle's Theorem can be applied, find all values of c in the interval. f(x)= x^2 - 2x -3/ (x +2) show all steps please

Answer 1

What is the interval given?

Answer 2

[-1,3]

Answer 3

The only value that would exclude the possibility for Rolle's theorem on this problem ix x=-2 (the value that would produce a divide by zero). You can use Rolle's Theorem because your interval is from -1 to 3 and does not include x=-2.

Answer 4

How do you solve this?

Answer 5

Check that f(-1)=f(3).

Answer 6

If it does, take the derivative of the function and set it equal to zero to find the x location where the function has a horizontal tangent... that is your c value.

Answer 7

how would you take the derivative of x^2 - 2x -3/ (x + 2)?

Answer 8

Quotient rule:\[\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}\]

Answer 9

show steps

Answer 10

\[f(x)=x ^{2}-2x-3\]\[f'(x)=2x-2\]\[g(x)=x+2\]\[g'(x)=1\]Now place these into the formula for the quotient rule above.

Answer 11

Let me know what you get.

Answer 12

where are getting g(x)?

Answer 13

The denominator.

Answer 14

Sorry, I re-used f(x)... most people look at that as the numerator for a quotient rule and g(x) as the denominator. I missed that f(x) was already used for your initial function. The idea is that you have a numerator function (usually f) and a denominator (usually g).

Answer 15

I hope that you have seen this before if you are working on Rolle's Theorem. Usually people do a whole chapter on derivative techniques before they tackle Rolle's.

Answer 16

I can re-state:\[\frac{d}{dx}[\frac{n}{d}]=\frac{n'd-nd'}{d ^{2}}\]\[n=x ^{2}-2x-3\]\[n'=2x-2\]\[d=x+2\]\[d'=1\]\[\frac{d}{dx}[\frac{n}{d}]=\frac{(2x-2)(x+2)-(x ^{2}-2x-3)(1)}{(x+2)^{2}}\]Now simplify.