Answer
(a) $f(-1) =-10\lt0$ and $f(0)= 2\gt0$.
(b) $f(2)= -4\lt0$ and $f(3)= 14\gt0$.
(c) $x=2.414$
Work Step by Step
(a) Given $f(x)=3x^3-8x^2+x+2$, we have $f(-1)=3(-1)^3-8(-1)^2+(-1)+2=-10\lt0$ and $f(0)=3(0)^3-8(0)^2+(0)+2=2\gt0$. Based on the intermediate theorem, there is a zero between -1 and 0.
(b) Similarly, we have $f(2)=3(2)^3-8(2)^2+(2)+2=-4\lt0$ and $f(3)=3(3)^3-8(3)^2+(3)+2=14\gt0$. Based on the intermediate theorem, there is a zero between 2 and 3.
(c) Graph the function as shown in the figure, we can identify the zero between 2 and 3 as $x=2.414$
