Using the Intermediate Value Theorem show there is a C, with -1

Question

Using the Intermediate Value Theorem show there is a C, with -1<c<1 where f(c)=0. For the equation F(x)=x^3+x^2-1

anonymous · Answer

plug in 1 and -1 in to the f(x) formula.. you should get (-1,-1) and (1,1)

anonymous · Answer

so as you are moving left to right from -1 to 1, you cross the x-axis since your v-values go from -1 to 1

zzr0ck3r · Answer

continuous*

anonymous · Answer

And to find the x value where f(x)=0, you set the whole equation equal to 0 and solve for x

anonymous · Answer

what do they mean by showing that there is a number C, with -1<c<1  such that f(x)=0

anonymous · Answer

When you go from -1 to 1 (on the y-axis), you cross 0 somewhere right?
so that's what they want you to find out... at what point do you cross 0 (or the x-axis)

anonymous · Answer

|dw:1360224875666:dw|    sorry it's sooo crooked lol

anonymous · Answer

even if it's not a straight line, you had to cross x-axis somewhere

anonymous · Answer

thanx i preciate it been needing help with this

anonymous · Answer

your welcome :D I didn't know this either until last week lol when i had a calc test and had to learn all this