Question

1. let $h(x)=\begin{cases}3x^{2}-4, &\text{if }xleq2\\5 + 4x, &\text{if }x>2end{cases}$. over the interval $0,4$, there is no value of $x$ such that $h(x)=10$. however, $h(0)<10$ and $h(4)>10$. explain why this does not contradict the intermediate value theorem.

Explanation:

The Intermediate Value Theorem requires a function to be continuous on the closed - interval. Here, the given function \(h(x)\) is a piece - wise function. It is discontinuous at \(x = 2\). Since the function is not continuous on the interval \([0,4]\), the Intermediate Value Theorem does not apply.

Answer:

The function \(h(x)\) is not continuous on \([0,4]\) so the Intermediate Value Theorem does not apply.