Using Rolle's Theorem Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers $C$ that satisfy the conclusion of Rolle's Theorem: $$f(x)=5-12x-3x^2 , [1,3]$$
Help with this would be so greatly appreciated :] I'm really not sure at all how to do this...
 A: If $f(x)=5-12x+3x^2$, then $f(1)=-4=f(3)$. Since $f$ is continuous on $[1,3]$ and differentiable on $(1,3)$, Rolle's theorem says there exists $c\in (1,3)$ with $f'(c)=0$. Indeed, $f'(x)=6x-12$ so $f'(2)=0$.
A: You need to show that


*

*The function $f$ is continuous on the interval $[1,3]$.

*The function $f$ is differentiable on the interval $(1,3)$.

*The function $f$ satisfies $f(1)=f(3)$.
Presumably, you are having trouble with the first two of these. The simplest solution would be to prove the following facts (it looks longer than it will take), or cite them, if they have been proven in class or in your textbook:


*

*Constant functions are continuous.

*Constant functions are differentiable on any open interval.

*The function $g(x)=x$ is continuous.

*The function $g(x)=x$ is differentiable on on any open interval.

*Sums and products of continuous functions are continuous.

*Sums and products of differentiable functions are differentiable.
Because a polynomial is a combination of sums and products of constant functions and the "identity" function $g$ -  for example,
$$f(x)=5-12x+3x^2=5-\bigg[12\cdot g(x)\bigg]+\bigg[3\cdot g(x)\cdot g(x)\bigg]$$
the above facts would then imply that $f$ (being a polynomial) satisfies these properties.
The conclusion of Rolle's theorem is a statement of the form

... so there exists some $c\in(1,3)$ such that $f'(c)=0$.

Thus, you are being instructed to find all $c\in (1,3)$ such that $f'(c)=0$. Do you know how to calculate $f'$, the derivative of this function $f$?
