Question 166553
Let's evaluate the left endpoint {{{x=-2}}}



{{{f(x)=4x^3-3x+3}}} Start with the given equation.



{{{f(-2)=4(-2)^3-3(-2)+3}}} Plug in {{{x=-2}}}.



{{{f(-2)=4(-8)-3(-2)+3}}} Cube {{{-2}}} to get {{{-8}}}.



{{{f(-2)=-32-3(-2)+3}}} Multiply {{{4}}} and {{{-8}}} to get {{{-32}}}.



{{{f(-2)=-32+6+3}}} Multiply {{{-3}}} and {{{-2}}} to get {{{6}}}.



{{{f(-2)=-23}}} Combine like terms.



-------------------------------------


Let's evaluate the right endpoint {{{x=-1}}}



{{{f(x)=4x^3-3x+3}}} Start with the given equation.



{{{f(-1)=4(-1)^3-3(-1)+3}}} Plug in {{{x=-1}}}.



{{{f(-1)=4(-1)-3(-1)+3}}} Cube {{{-1}}} to get {{{-1}}}.



{{{f(-1)=-4-3(-1)+3}}} Multiply {{{4}}} and {{{-1}}} to get {{{-4}}}.



{{{f(-1)=-4+3+3}}} Multiply {{{-3}}} and {{{-1}}} to get {{{3}}}.



{{{f(-1)=2}}} Combine like terms.



So as x changes from -2 to -1, f(x) (or y) changes from -23 to 2 which means that the graph MUST have crossed over the x-axis somewhere in between x=-2 and x=-1. So this shows that there is a zero between x=-2 and x=-1