This is a sampling of basic problems that you might expect to have on the test. Do not expect that the test will contain such problems, or will be limited to this difficulty of question. In fact, expect about 50-60% of the points to be of this type, and the other 40% or so to be more difficult. That is, in order to pass you should be able to do all such problems accurately. Okay, enought said.

plot the equation $x^2 = y^3$

ANSWER: Simply plot the function $y = x^{2/3}$ with your calculator or by plotting a table of points

sketch the function $g(x) = 300 + \sqrt{300 + x}$

ANSWER: This is a square root graph shifted up 300 units and left 300 units.

Sketch two periods of the graph of $f(x) = 3 + 4 \sin(5x + 6)$

ANSWER: This is a sine curve with period $T = 2\pi/5$ shifted left by $c/b=6/5$ with an amplitude of $A=4$ and shifted up by 3.

Compute the limits.

$$\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(4x)},\quad \lim_{x \rightarrow 0} \frac{\sqrt{4+x} - 2}{x},\quad \lim_{x \rightarrow 0} \frac{x^3 + x^2 + x}{x^2 + x},\quad$$

ANSWER:

$$\lim_{x \rightarrow 0} \frac{\sin(3x)}{\sin(4x)} = \lim_{x \rightarrow 0} \frac{\sin(3x)}{3x}\frac{4x}{\sin(4x)}\frac{3}{4}=3/4$$

$$\lim_{x \rightarrow 0} \frac{\sqrt{4+x} - 2}{x} = \lim_{x \rightarrow 0} \frac{(4+x) - 4}{x(\sqrt{4+x} + 2)} = 1/4$$

$$\lim_{x \rightarrow 0} \frac{x^3 + x^2 + x}{x^2 + x} = \lim_{x \rightarrow 0} \frac{x(x^2 + x + 1}{x(x+1)} = \frac{0^2 + 0 +1 }{0+1} = 1$$

Compute $\lim_{x \rightarrow 0-} |x| /x$.

ANSWER: For $x < 0$ $|x|/x = -1$ so limit is -1.

Where is the function $f(x) = \sin(x)/(x(x+2))$ discontinuous. Are any of these points removable singularities?

ANSWER: The function is continuous everywhere except when the denominator is 0 as both the top and bottom functions are continuous everywhere. The denominator is 0 at $x = 0$ and $x=2$. The value $x=0$ is a removable singularity as the funciton $f(x)$ has a limit of 1/3 there. At $x=2$ there is a vertical asymptote.

Sketch the graph of $f(x) = (x+5)^2$, Sketch the tangent line at $x = -2$ and find the equation of the tangent line.

ANSWER: THe slope at $x=-2$ is $f'(-2) = 2(-2+5)^1 = -6$ . A point is $(-2,9)$ so the equation is $y = -6(x+2) -9$

Find the derivatives of the following functions:

$$f(x) = x^2 - 16x,\quad g(x) = x^2 \sin(x),\quad h(x) = x^2 / \sin(x),\quad i(x) = \sin(x^2 - 16x)$$

ANSWER:

$$2x - 16,\quad (2x)\sin(x) + x^2(\cos(x)),\quad \frac{(2x)\sin(x) - x^2(\cos(x))}{\sin^2(x)},\quad \cos(x^2 - 16x)(2x - 16)$$

Find the second derivative of $f(x) = x^16 - \sin(x) + \sqrt{x}$.

ANSWER:

$$f'(x) = 16x^15 -\cos(x) + (1/2) x^{-1/2},\quad f''(x) = (16)(15)x^14 - (-\sin(x)) + (1/2)(-1/2)x^{-3/2}$$