a relative maximum at \( \displaystyle x = \frac{1}{2}\)
a horizontal asymptote at \( y = 0\)
a vertical asymptote at \( x = 0\)
I only
I and II
I and III
II and III
I, II and III
Solution
D
Given the piecewise function defined as \( \displaystyle
f(x) =
\begin{cases}
x^2 + 2, & \text{if } x < 1 \\
2x + 1, & \text{if }x \geq 1
\end{cases}\), which of the following is true?
\( f(x)\) is not continuous at \(x = 1\).
\( f(x)\) is continuous but not idfferentiable at \( x = 1\).
\( f(x)\) is differentiable but not continuous at \( x = 1\).
\( f(x)\) is continuous and differentiable at \( x = 1\).
\( f(2) = 6\)
Solution
D
\( \displaystyle \int_1^e x \ln{x} \, dx =\)
\( \displaystyle \frac{e^2+ 1}{4}\)
\(2e^2 - 1\)
\( \displaystyle \frac{e^2 - e - 1}{2}\)
\(2e^2 + 1\)
undefined
Solution
A
Write an integral that represents the length of one arch of \(y = \sin{ x}\).
For what values of \(a\) and \(c\) is the piecewise function defined by
\( \displaystyle f(x) =
\begin{cases}
x + c, & \text{if } x > 2 \\
ax^2, & \text{if }x \leq 2
\end{cases}\) differentiable at x = 2?
\( \displaystyle a = \frac{1}{2}, c= 0\)
\( \displaystyle a = \frac{1}{4}, c = -1\)
\(a = 1, c = 6\)
\(a = 0, c = -2\)
no solution
Solution
B
The slope field below depicts a certain differential equation. Which of the following choices could be a solution to that equation?
\(y = \ln{x}\)
\(y = e^{-x}\)
\(y = \sin{x}\)
\(y = e^x\)
\(y = x^{2\sin{ x}}\)
Solution
B
\( \displaystyle h(x) = \frac{f(x)}{\left( g(x) \right)^2}\). If \( \displaystyle f'(x) = g(x), g'(x) = \frac{1}{f(x)}, g(x) > 0\) for all real \(x\), and \(f(x) \neq 0\) for all real \(x\), then \(h'(x) =\)
Given the graph of \(y = f(x)\) shown below, if \( \displaystyle g(x) = f\left(\frac{1}{x}\right) \), find the value of \( \displaystyle \lim_{x \to 0^+} g(x)\)
-1
0
1
\(\infty\)
does not exist
Solution
C
\(y = \ln{ (\cos^2{x})}. y' =\)
\(-2 \tan{ x}\)
\(\sec^2 {x}\)
\(2 \sec {x}\)
\(2 \tan{ x}\)
\(-2 \sin{x} \cos{ x}\)
Solution
A
Write the equation of the line perpendicular to the tangent of the curve represented by the equation \(y = e^{x+1}\) at \(x = 0\)
\( \displaystyle y = -\frac{x}{e}\)
\( \displaystyle y = -\frac{x}{e} + e\)
\(y = ex + e\)
\( \displaystyle y = \frac{x}{e} + e\)
\(y = ex \)
Solution
B
Find the area bounded by the graph of \( \displaystyle y = \frac{x}{x^2 - 1}\) and the x-axis on the interval \( \displaystyle \left[-\frac{1}{2}, \frac{1}{2} \right]\).