1.
Question 1 Evaluate \(\lim_{x\to 0}\frac{\sin 4x}{\sin 5x}\).
Solution: Use the standard limit \(\lim_{u\to 0}\frac{\sin u}{u}=1\). Rewrite \(\frac{\sin 4x}{\sin 5x}=\frac{4}{5}\cdot\frac{\sin 4x}{4x}\cdot\frac{5x}{\sin 5x}\). As \(x\to0\), \(\frac{\sin 4x}{4x}\to1\) and \(\frac{5x}{\sin 5x}\to1\). Therefore the limit is \(\frac{4}{5}\cdot1\cdot1=\frac{4}{5}\). Answer: \(\frac{4}{5}\)
2.
Question 2 Evaluate \(\lim_{x\to 0}\frac{\sin 6x}{\sin 3x}\).
Solution: Use the standard limit \(\lim_{u\to0}\frac{\sin u}{u}=1\). Rewrite \(\frac{\sin 6x}{\sin 3x}=\frac{6}{3}\cdot\frac{\sin 6x}{6x}\cdot\frac{3x}{\sin 3x}\). As \(x\to0\), both limit factors approach 1, so the limit is \(\frac{6}{3}\cdot1\cdot1=2\). Answer: 2
3.
Question 3 Where is \(f(x)\) not continuous?
Solution: From the graph, there are vertical asymptotes at \(x=-2\) and \(x=2\). A function is not continuous where it has a vertical asymptote. Therefore, \(f(x)\) is not continuous at \(x=-2\) and \(x=2\). Answer: -2 and 2
4.
Question 4 The Intermediate Value Theorem can be used to prove that the equation \(3x^2-x-2=0\) has at least one solution on the interval \([0,2]\).
Solution: Let \(f(x)=3x^2-x-2\). Since \(f(x)\) is a polynomial, it is continuous everywhere, including on \([0,2]\). Evaluate the endpoints: \(f(0)=3(0)^2-0-2=-2\) and \(f(2)=3(2)^2-2-2=8\). Since \(f(0)0\), the function changes sign on \([0,2]\). By the Intermediate Value Theorem, there is at least one value \(c\in[0,2]\) such that \(f(c)=0\). Answer: Ture
5.
Question 5 What is the instantaneous rate of change of \(y=\sqrt{x}\) at \(x=2\)?
Solution: The instantaneous rate of change is the derivative. Since \(y=\sqrt{x}=x^{1/2}\), \(y'=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\). At \(x=2\), \(y'=\frac{1}{2\sqrt{2}}\). Answer: \(\frac{1}{2\sqrt{2}}\)
6.
Question 6 What is the instantaneous rate of change of \(y=\frac{1}{x}\) at \(x=2\)?
Solution: The instantaneous rate of change is the derivative. Since \(y=\frac{1}{x}=x^{-1}\), \(y'=-x^{-2}=-\frac{1}{x^2}\). At \(x=2\), \(y'=-\frac{1}{2^2}=-\frac{1}{4}\). Answer: \(-\frac{1}{4}\)
7.
Question 7 The average rate of change of \(y\) with respect to \(x\) over the interval \([0,3]\) for the function \(y=3x+2\) is 9.
Solution: The average rate of change over \([0,3]\) is \(\frac{y(3)-y(0)}{3-0}\). For \(y=3x+2\), \(y(0)=2\) and \(y(3)=11\). Therefore, the average rate of change is \(\frac{11-2}{3-0}=\frac{9}{3}=3\), not 9. The statement is false. Answer: False
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