let (g) and (h) be the functions defined by (g(x)=sinleft(\frac{pi}{2}(x + 2)\right)+3) and…

let (g) and (h) be the functions defined by (g(x)=sinleft(\frac{pi}{2}(x + 2)\right)+3) and (h(x)=-\frac{1}{4}x^{3}-\frac{3}{2}x^{2}-\frac{9}{4}x + 3). if (f) is a function that satisfies (g(x)leq f(x)leq h(x)) for (-2 < x < 0), what is (lim_{x\rightarrow - 1}f(x))?\na 3\nb 3.5\nc 4\nd the limit cannot be determined from the information given.

let (g) and (h) be the functions defined by (g(x)=sinleft(\frac{pi}{2}(x + 2)\right)+3) and (h(x)=-\frac{1}{4}x^{3}-\frac{3}{2}x^{2}-\frac{9}{4}x + 3). if (f) is a function that satisfies (g(x)leq f(x)leq h(x)) for (-2 < x < 0), what is (lim_{x\rightarrow - 1}f(x))?\na 3\nb 3.5\nc 4\nd the limit cannot be determined from the information given.

Answer

Explanation:

Step1: Apply 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\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L), then (\lim_{x\rightarrow a}f(x)=L). Here (a=-1), and (g(x)\leq f(x)\leq h(x)) for (- 2<x<0) which contains (x = - 1).

Step2: Calculate (\lim_{x\rightarrow - 1}g(x))

Substitute (x=-1) into (g(x)=\sin\left(\frac{\pi}{2}(x + 2)\right)+3). First, find the value inside the sine function: (\frac{\pi}{2}(-1 + 2)=\frac{\pi}{2}). Then (g(-1)=\sin\left(\frac{\pi}{2}\right)+3). Since (\sin\left(\frac{\pi}{2}\right)=1), (g(-1)=1 + 3=4).

Step3: Calculate (\lim_{x\rightarrow - 1}h(x))

Substitute (x=-1) into (h(x)=-\frac{1}{4}x^{3}-\frac{3}{2}x^{2}-\frac{9}{4}x + 3). [ \begin{align*} h(-1)&=-\frac{1}{4}(-1)^{3}-\frac{3}{2}(-1)^{2}-\frac{9}{4}(-1)+3\ &=\frac{1}{4}-\frac{3}{2}+\frac{9}{4}+3\ &=\frac{1 - 6+9 + 12}{4}\ &=\frac{16}{4}\ &=4 \end{align*} ] Since (\lim_{x\rightarrow - 1}g(x)=\lim_{x\rightarrow - 1}h(x)=4), by the Squeeze Theorem, (\lim_{x\rightarrow - 1}f(x)=4).

Answer:

C. 4