Answer
(a) $f(2) =44\gt0$ and $f(3) =-15\lt0$.
(b) $f(7) =-31\lt0$ and $f(8) =140\gt0$.
(c) $x=7.236$
Work Step by Step
(a) Given $f(x)=4x^3-37x^2+50x+60$, we have $f(2)=4(2)^3-37(2)^2+50(2)+60=44\gt0$ and $f(3)=4(3)^3-37(3)^2+50(3)+60=-15\lt0$. Based on the intermediate theorem, there is a zero between 2 and 3.
(b) Similarly, we have $f(7)=4(7)^3-37(7)^2+50(7)+60=-31\lt0$ and $f(8)=4(8)^3-37(8)^2+50(8)+60=140\gt0$. Based on the intermediate theorem, there is a zero between 7 and 8.
(c) Graph the function as shown in the figure, we can identify the zero between 7 and 8 as $x=7.236$
