An equation of the line tangent to the graph of \(y = \dfrac{2 x + 3}{3 x - 2}\) at the point \((1, 5)\) is

If \(f(x) = (x^2 - 2 x - 1)^{\dfrac{2}{3}}\), then \(f'(0)\) is

If \(f(x) = e^{3 \ln(x^2)}\), then \(f'(x) =\)

If \(x^3 + 3 x y + 2 y^3 = 17\), then in terms of \(x\) and \(y\), \(\dfrac{d y}{d x} =\)

\(\dfrac{d}{d x}(2^x) =\)

[Calculator] A particle moves along a line so that at time \(t\), where \(0 \leq t \leq \pi\), its position is given by \(s(t) = -4 \cos t - \dfrac{t^2}{2} + 10\). What is the velocity of the particle when its acceleration is zero?

If \(f\) is a differentiable function, then \(f'(a)\) is given by which of the following? I. \(\operatorname*{lim}\limits_{h \rightarrow 0} \dfrac{f(a + h) - f(a)}{h}\) II. \(\operatorname*{lim}\limits_{x \rightarrow a} \dfrac{f(x) - f(a)}{x - a}\) III. \(\operatorname*{lim}\limits_{x \rightarrow a} \dfrac{f(x + h) - f(x)}{h}\)

\(\displaystyle\int_{1}^{2} (4 x^3 - 6 x) d x =\)

\(\dfrac{1}{2} \int e^{\dfrac{t}{2}} d t =\)

A bug begins to crawl up a vertical wire at time \(t = 0\). The velocity \(v\) of the bug at time \(t\), \(0 \leq t \leq 8\), is given by a piecewise-linear graph: \(v\) rises linearly from \(0\) at \(t=0\) to \(3\) at \(t=2\), stays at \(3\) for \(2 \leq t \leq 4\), decreases linearly from \(3\) to \(0\) at \(t=6\), drops to \(-1\) at \(t=7\), and rises back to \(0\) at \(t=8\). At what value of \(t\) does the bug change direction?

Using the velocity graph from the previous problem (rises linearly from \(0\) to \(3\) on \([0,2]\), constant \(3\) on \([2,4]\), decreases linearly to \(0\) on \([4,6]\), dips to \(-1\) at \(t=7\), returns to \(0\) at \(t=8\)), what is the total distance the bug traveled from \(t = 0\) to \(t = 8\)?

\(\displaystyle\int_{0}^{\dfrac{\pi}{4}} \dfrac{e^{\tan x}}{\cos^2 x} d x\) is

\(\displaystyle\int_{0}^{1} \sqrt{x} (x + 1) d x\)

[Calculator] At time \(t \geq 0\), the acceleration of a particle moving on the x-axis is \(a(t) = t + \sin t\). At \(t = 0\), the velocity of the particle is \(-2\). For what value of \(t\) will the velocity of the particle be zero?

A table of values for a continuous function \(f\): \(f(0)=3\), \(f(0.5)=3\), \(f(1.0)=5\), \(f(1.5)=8\), \(f(2.0)=13\). If four equal subintervals of \([0, 2]\) are used, which of the following is the midpoint approximation of \(\displaystyle\int_{0}^{2} f(x) d x\)?

The data for the acceleration \(a(t)\) of a car from \(0\) to \(6\) seconds are given: \(a(0)=5\), \(a(2)=2\), \(a(4)=8\), \(a(6)=3\) (ft/sec\(^2\)). If the velocity at \(t = 0\) is \(11\) feet per second, the approximate value of the velocity at \(t = 6\), computed using a left-hand Riemann sum with three subintervals of equal length, is

\(\int \dfrac{3 x^2}{\sqrt{x^3 + 1}} d x =\)

\(\int (x^2 + 1)^2 d x =\)