1.
Question 1. Where is \(f(x)=\frac{x-3}{x^2+3x-18}\) discontinuous?
A rational function is discontinuous where its denominator equals zero. Set \(x^2+3x-18=0\). Factor: \((x+6)(x-3)=0\). Therefore, \(x=-6\) or \(x=3\). The function is discontinuous at \(-6\) and \(3\). Answer: B
2.
Question 2. Where is \(f(x)\) not continuous?
From the graph, there are vertical asymptotes at \(x=-2\) and \(x=2\). A function is not continuous at vertical asymptotes. Therefore, \(f(x)\) is not continuous at \(-2\) and \(2\). Answer: A
3.
Question 3. 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 \([0,2]\).
Let \(f(x)=3x^2-x-2\). Since \(f(x)\) is a polynomial, it is continuous 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: A
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Question 4. \(f(x)=\begin{cases}x^2-2, & x\le 3\\ x+4, & x>3\end{cases}\) is continuous everywhere.
Both pieces are continuous on their own intervals, so only \(x=3\) needs to be checked. The left-hand value is \(3^2-2=7\). The right-hand value approaches \(3+4=7\). Also, \(f(3)=3^2-2=7\). Since the left-hand limit, right-hand limit, and function value are all equal, the function is continuous at \(x=3\). Therefore, it is continuous everywhere. Answer: A
5.
Question 5. The position of a particle on the x-axis is given by \(x(t)=t^2-4t+3\). At what time is the particle’s velocity zero?
Velocity is the derivative of position. Since \(x(t)=t^2-4t+3\), we have \(v(t)=x'(t)=2t-4\). Set the velocity equal to zero: \(2t-4=0\). Then \(2t=4\), so \(t=2\). Answer: D
6.
Question 6. What is the instantaneous rate of change of \(y=\frac{1}{x}\) at \(x=2\)?
The instantaneous rate of change is the derivative. For \(y=\frac{1}{x}=x^{-1}\), the derivative is \(y'=-x^{-2}=-\frac{1}{x^2}\). At \(x=2\), \(y'=-\frac{1}{2^2}=-\frac{1}{4}\). Answer: E
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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 \(9\).
The average rate of change is \(\frac{y_2-y_1}{x_2-x_1}\). For \(y=3x+2\), \(y(0)=3(0)+2=2\), and \(y(3)=3(3)+2=11\). Therefore, the average rate of change is \(\frac{11-2}{3-0}=\frac{9}{3}=3\), not \(9\). The statement is false. Answer: A
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