Question

closed interval 0, 4, where h(0) = 2 and h(4) = -2. which of the following is guaranteed by the intermediate value theorem? choose 1 answer: a h(c) = -1 for at least one c between 0 and 4 b h(c) = 3 for at least one c between -2 and 2 c h(c) = -1 for at least one c between -2 and 2 d h(c) = 3 for at least one c between 0 and 4

Explanation:

Step1: Recall Intermediate - Value Theorem

If a function $h(x)$ is continuous on a closed interval $[a,b]$, and $k$ is a number between $h(a)$ and $h(b)$, then there exists at least one number $c$ in the interval $(a,b)$ such that $h(c)=k$. Here, $a = 0$, $b = 4$, $h(0)=2$ and $h(4)=-2$.

Step2: Check the range of values

The values of the function $h(x)$ on the interval $[0,4]$ range from $-2$ to $2$ (since $h(0) = 2$ and $h(4)=-2$). We need to find a value $k$ such that $-2\leq k\leq2$ and then there will be a $c\in(0,4)$ with $h(c)=k$.

Step3: Analyze each option

• Option A: Since $-2<-1 < 2$, by the Intermediate - Value Theorem, there exists at least one $c$ between $0$ and $4$ such that $h(c)=-1$.
• Option B: The interval for $c$ is wrong. The theorem gives a $c$ in the domain interval $[0,4]$, not in the range interval $[-2,2]$. Also, $3$ is outside the range $[-2,2]$.
• Option C: The interval for $c$ is wrong. The theorem gives a $c$ in the domain interval $[0,4]$, not in the range interval $[-2,2]$.
• Option D: $3$ is outside the range of values of $h(x)$ on $[0,4]$ (the range is $[-2,2]$).

Answer:

A. $h(c)=-1$ for at least one $c$ between $0$ and $4$