Question

How do you use the intermediate value theorem to explain why `f(x)=1/16x^4-x^3+3` has a zero in the interval [1,2]?

Answer 1

The function is continuous and `f(x)=0` exists in the interval [1,2]

Explanation:

The function `f(x)` is continuous in the interval [1,2] as it is a polynomial function.
Therefore we calculate
`f(1)=1/16-1+3=33/16`
and `f(2)=1-8+3=-4`
Therefore as `-4<=f(x)<=33/16`
As one value is above zero and the other below zero
`f(x)=0` exists in the interval [1,2]

graph{x^4/16-x^3+3 [-5.94, 6.55, -3.34, 2.9]}