Question

let g(x) and h(x) be defined by ( g(x) = -x^2 - 3x ) and ( h(x) = -11x + 16 ). if ( f(x) ) is a function such that ( g(x) leq f(x) leq h(x) ), find ( limlimits_{x \to 4} f(x) ), or state that it cannot be determined.

Explanation:

Step1: Find the limit of \( g(x) \) as \( x \to 4 \)

We have \( g(x)=-x^{2}-3x \). To find \( \lim_{x \to 4}g(x) \), we substitute \( x = 4 \) into \( g(x) \):
\( \lim_{x \to 4}(-x^{2}-3x)=-(4)^{2}-3\times4=-16 - 12=-28 \)

Step2: Find the limit of \( h(x) \) as \( x \to 4 \)

We have \( h(x)=-11x + 16 \). To find \( \lim_{x \to 4}h(x) \), we substitute \( x = 4 \) into \( h(x) \):
\( \lim_{x \to 4}(-11x + 16)=-11\times4+16=-44 + 16=-28 \)

Step3: Apply the Squeeze Theorem

The Squeeze Theorem states that if \( g(x)\leq f(x)\leq h(x) \) for all \( x \) near \( a \) (except possibly at \( a \)) and \( \lim_{x \to a}g(x)=\lim_{x \to a}h(x)=L \), then \( \lim_{x \to a}f(x)=L \).
Here, \( a = 4 \), \( \lim_{x \to 4}g(x)=\lim_{x \to 4}h(x)=-28 \), and \( g(x)\leq f(x)\leq h(x) \). So by the Squeeze Theorem, \( \lim_{x \to 4}f(x)=-28 \)

Answer:

\( -28 \)