Answer
a. $f(x)=x(x+1)(x-1)(x+2)(x-2)$
b. See graph and explanations.
c. Yes.
Work Step by Step
a. Given zeros $x=0,\pm1,\pm2$ for a polynomial, we can write the function as $f(x)=x(x+1)(x-1)(x+2)(x-2)=x(x^2-1)(x^2-4)=x(x^4-5x^2+4)=x^5-5x^3+4x$
b. We can find $f'(x)=5x^4-15x^2+4$ and we can graph both $f$ and $f'$ as shown in the figure. We can see that there is a zero in $f'$ between any two zeros of $f$, which agrees with the Rolle's Theorem.
c. If $g(x)=sin(x)$, we have $g'(x)=cos(x)$; the zeros for $g(x)$ are $x=k\pi$ and zeros for $g'(x)$ are $x=k\pi+\pi/2$, and we can see that there is a zero in $g'$ between any two zeros in $g$ which also agrees with Rolle's Theorem.
