Question

let $g$ and $h$ be the functions defined by $g(x)=-x^{2}-2x + 3$ and $h(x)=\frac{1}{2}x^{2}+x+\frac{13}{2}$. if $f$ is a function that satisfies $g(x)leq f(x)leq h(x)$ for all $x$, what is $lim_{x
ightarrow - 1}f(x)$?
a 4
b 5
c 6
d the limit cannot be determined from the information given.

Explanation:

Step1: Recall the Squeeze - Theorem

The Squeeze - Theorem states that if \(g(x)\leq f(x)\leq h(x)\) for all \(x\) in an open interval containing \(a\) (except possibly at \(x = a\)) and \(\lim_{x
ightarrow a}g(x)=\lim_{x
ightarrow a}h(x)=L\), then \(\lim_{x
ightarrow a}f(x)=L\). Here \(a=- 1\).

Step2: Calculate \(\lim_{x

ightarrow - 1}g(x)\)
Substitute \(x=-1\) into \(g(x)=-x^{2}-2x + 3\).
\[

$$\begin{align*} \lim_{x ightarrow - 1}g(x)&=-(-1)^{2}-2\times(-1)+3\\ &=-1 + 2+3\\ &=4 \end{align*}$$

\]

Step3: Calculate \(\lim_{x

ightarrow - 1}h(x)\)
Substitute \(x = - 1\) into \(h(x)=\frac{1}{2}x^{2}+x+\frac{13}{2}\).
\[

$$\begin{align*} \lim_{x ightarrow - 1}h(x)&=\frac{1}{2}\times(-1)^{2}+(-1)+\frac{13}{2}\\ &=\frac{1}{2}-1+\frac{13}{2}\\ &=\frac{1 - 2+13}{2}\\ &=\frac{12}{2}\\ &=4 \end{align*}$$

\]

Step4: Apply the Squeeze - Theorem

Since \(g(x)\leq f(x)\leq h(x)\) for all \(x\) and \(\lim_{x
ightarrow - 1}g(x)=\lim_{x
ightarrow - 1}h(x)=4\), by the Squeeze - Theorem, \(\lim_{x
ightarrow - 1}f(x)=4\).

Answer:

A. 4