f(x)=x^12-x^8+x^2016 show that there is some real number 0<x<1, such that f(x)=0

Question

You're supposed to use intermediate value theorem, but I'm having trouble because f(0)=0 and I can't find another number to equal 0

Answer 1

f(½) = something negative; essentially 1/4096 - 1/256 (the third term is negligible)

f(1) = 1

 

By IVT, there's some zero on (½,1) because this is a polynomial & it's therefore continuous.