# 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]?

Nov 3, 2016

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

#### Explanation:

The function $f \left(x\right)$ is continuous in the interval [1,2] as it is a polynomial function.
Therefore we calculate
$f \left(1\right) = \frac{1}{16} - 1 + 3 = \frac{33}{16}$
and $f \left(2\right) = 1 - 8 + 3 = - 4$
Therefore as $- 4 \le f \left(x\right) \le \frac{33}{16}$
As one value is above zero and the other below zero
$f \left(x\right) = 0$ exists in the interval [1,2]

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