Question

let g(x) and h(x) be defined by ( g(x) = -2x^2 + 4x ) and ( h(x) = 16x + 18 ). if ( f(x) ) is a function such that ( g(x) leq f(x) leq h(x) ), find ( limlimits_{x \to -3} f(x) ), or state that it cannot be determined.

Explanation:

Step1: Find limit of \( g(x) \) as \( x \to -3 \)

We have \( g(x) = -2x^2 + 4x \). Substitute \( x = -3 \) into \( g(x) \):
\( g(-3)=-2(-3)^2 + 4(-3)=-2\times9 - 12=-18 - 12=-30 \)
So, \( \lim_{x\to -3}g(x)=-30 \) (since \( g(x) \) is a polynomial, its limit at any point is its value at that point).

Step2: Find limit of \( h(x) \) as \( x \to -3 \)

We have \( h(x)=16x + 18 \). Substitute \( x = -3 \) into \( h(x) \):
\( h(-3)=16(-3)+18=-48 + 18=-30 \)
So, \( \lim_{x\to -3}h(x)=-30 \) (since \( h(x) \) is a polynomial, its limit at any point is its value at that point).

Step3: Apply 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, \( a = -3 \), \( \lim_{x\to -3}g(x)=\lim_{x\to -3}h(x)=-30 \), and \( g(x)\leq f(x)\leq h(x) \). So by Squeeze Theorem, \( \lim_{x\to -3}f(x)=-30 \).

Answer:

\( -30 \)