Rolle's Theorem

[Violagirl]

Hi, I am stuck to a certain point on this problem:

1. Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(c)=0.

f(x)=(x-1) (x-2) (x-3), [1,3]

I know Rolle's Theorem can be applied since f is continuous along the interval of [1,3] and differentiable along that interval as well.

f(x) = x[sup:cghctzop]3[/sup:cghctzop]-6x[sup:cghctzop]2[/sup:cghctzop]+11x-6
f'(x) = 3x[sup:cghctzop]2[/sup:cghctzop]-12x+11

I'm stuck after that, how would you find the values of c? Thanks!

 

[Deleted member 4993]

Hi, I am stuck to a certain point on this problem:

1. Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If Rolle's Theorem can be applied, find all values of c in the open interval (a,b) such that f'(c)=0.

f(x)=(x-1) (x-2) (x-3), [1,3]

I know Rolle's Theorem can be applied since f is continuous along the interval of [1,3] and differentiable along that interval as well.

f(x) = x[sup:bmcxsu99]3[/sup:bmcxsu99]-6x[sup:bmcxsu99]2[/sup:bmcxsu99]+11x-6
f'(x) = 3x[sup:bmcxsu99]2[/sup:bmcxsu99]-12x+11

I'm stuck after that, how would you find the values of c? Thanks!

find all values of c in the open interval (a,b) such that f'(c)=0.

Read your problem statement carefully.

 

[BigGlenntheHeavy]

\(\displaystyle Rolle's \ Theorem\)

\(\displaystyle Let \ f \ be \ differential \ on \ (a,b) \ and \ continuous \ on \ [a,b].\)

\(\displaystyle If \ f(a) \ = \ f(b) \ = \ 0, \ then \ there \ is \ at \ least \ one \ point \ c \ in \ (a,b) \ where \ f \ ' \ (c) \ = \ 0.\)

\(\displaystyle f(x) \ = \ (x-1)(x-2)(x-3), \ f(1) \ = \ f(3) \ = \ 0, \ hence \ f \ ' \ (c) \ = \ 0.\)

\(\displaystyle f \ ' \ (x) \ = \ 3x^{2}-12x+11, \ f \ ' \ (c) \ = \ 3c^{2}-12c+11 \ = \ 0, \ solve \ for \ c \ and \ you \ are \ done.\)

\(\displaystyle See \ graph.\)

[attachment=0:ioe5airh]pac.jpg[/attachment:ioe5airh]

 

[Violagirl]

Thank you both very much, I figured it out.