Exercises - Review of Some Ideas from Precalculus

  1. Factor $xy^2 + y^2 - 4x -4$

    $(y-2)(y+2)(x+1)$

  2. Simplify $\displaystyle{\frac{f(x)}{g(x)}}$, where $\displaystyle{f(x) = \frac{x-2}{x} + \frac{x}{x+2} \textrm{ and } g(x) = \frac{x-2}{x+2} + \frac{1}{x}}$

    $\displaystyle{\frac{f(x)}{g(x)} = \frac{2x^2-4}{x^2-x+2} \quad \textrm{where } x \ne 0,-2}$

  3. Solve $x^3 + 3x^2 = x + 3$

    $x = -3,-1,1$

  4. Solve $\displaystyle{\frac{x^2}{x^2-1} - \frac{2}{x+1} = 1}$

    $x = 3/2$

  5. Solve $x+1 = \sqrt{x+7}$

    $x=2$

  6. Find an equation of the line $\ell_1$ through $(-3,2)$ and $(2,-1)$, and then find the equation of the line $\ell_2$ passing through $(-1,-3)$ and perpendicular to $\ell_1$.

    Using point-slope form in conjunction with the first point given yields the equation $y-2 = \frac{-3}{5}(x+3)$ for $\ell_1$. Finding the perpendicular slope by reciprocating and negating the slope of $\ell_1$, and then using point-slope form in conjunction with $(-1,-3)$ yields $y+3 = \frac{5}{3}(x+1)$ as an equation for $\ell_2$.

  7. Find the center and radius of the circle described by $x^2 + y^2 - 4x + 2y + 1 = 0$

    center $(2,-1)$; radius $2$

  8. Find $\sec \theta$ if $\csc \theta = -3$ and $\cot \theta \lt 0$

    $\displaystyle{\frac{3 \sqrt{2}}{4}}$

  9. Solve $2\cos^2 x + \sin x - 1 = 0$

    $\displaystyle{x = \frac{7\pi}{6} \pm 2\pi n, \frac{11\pi}{6} \pm 2\pi n, \frac{\pi}{2} \pm 2\pi n, \textrm{ for } n \in \mathbb{N}}$

  10. Find $\displaystyle{\textrm{Arcsin} \left( \sin \frac{3\pi}{4} \right)}$

    $\displaystyle{\frac{\pi}{4}}$

  11. Find $\displaystyle{\tan \left( 2\textrm{Arctan}(-\sqrt{3}) \right)}$

    $\sqrt{3}$

  12. Solve $2 \log_9 x = 1$

    $x = 3$

  13. Solve $\displaystyle{4^{-x+1} = \frac{1}{32}}$

    $\displaystyle{x = \frac{7}{2}}$

  14. Solve $2^x + 2^{-x} = 2$

    $x = 0$

  15. Solve $\displaystyle{\log_3 (7-x) = \log_3 (1-x) + 1}$

    $x = -2$

  16. Solve $\displaystyle{\frac{\log_3 16}{2 \log_3 x} = 2}$

    $x = 2$

  17. Define what it means for a function to be "even" or "odd." Provide an example of each.

    A function $f$ is even when $f(-x) = f(x)$ for all $x$ in its domain. Visually, one can fold even functions along the $y$-axis and the two halves formed will line up perfectly. Examples include $x^2 + 3, \cos x, |x|$, and many others.

    A function $g$ is odd when $g(-x) = -g(x)$ for all $x$ in its domain. Visually, odd functions are rotationally symmetric about the origin. That is to say, one can rotate the right half of the function's graph 180 degrees about the origin and it will line up perfectly with the left half of the graph. Examples of odd functions include $x^3 -x, \sin x, 1/x$, and many others.

    Notably, functions need not be one or the other (i.e., either even or odd). For example, $x^2 + x$, $\ln x$, and $e^x$ are each neither even nor odd. Alternatively, one function $h(x) = 0$ is both even and odd.

  18. A small rectangular pen containing $10$ square meters is to be fenced in. The front, to be made of stone, will cost $\$7$ per meter, while each of the other three wooden sides will cost only $\$3$ per meter. Supposing the front of the pen is $x$ meters long and the sides are each $y$ meters long, find the cost of the fence in terms of these two dimensions. Can you find the cost in terms of only the length of the front of the pen? What happens to the cost when $x$ increases?

    $C = 10x + 6y$. As $xy = 10$, we can solve to find $y = 10/x$ which allows us to express the cost as a function of only one variable: $C(x) = 10x + 60/x$ where $x \gt 0$. For very small lengths $x$, the cost is large as the side lengths must be exceedingly long to accomplish the desired area of $10$ square meters. However, as $x$ increases, the first term of $C(x)$ increases, while the second diminishes -- until the cost is driven almost solely by the first term (the second gets very close to zero for large $x$). Again, this is expected as the side lengths need not be long (and are consequently cheap) when the front length is large enough.

  19. A box with a square base and open top must have a volume of $32,000\, \textrm{cm}^3$. Find an expression for the amount of material required to make the box in terms of its dimensions. Can you express this amount of material in terms of only the base length $x$ (in cm)?

    Supposing $S$ is the amount of material, $x$ is the base length (in cm), and $y$ is the height of the box (in cm), we find $S = x^2 + 4xy$. Taking advantage of the volume being $x^2 y = 32000$ we find $y = 32000/x^2$ which allows us to express $S$ as function of one variable: $$S(x) = x^2 + \frac{128000}{x} \quad \textrm{ where } \quad x \gt 0$$

  20. A box with a square base and an open top is to be made using $1200\, \textrm{cm}^2$ of material. Let $x$ be the length (in cm) of a side of the base. Find the volume of the box as a function of $x$, noting any restrictions on the values of $x$ forced upon one given that the volume of an actual box must always be positive.

    $V(x) = (x/4)(1200-x^2)$ where $0 \lt x \lt \sqrt{1200}$

  21. A right circular cylinder is inscribed in a sphere of radius $r$. Find the volume of the cylinder as a function of the height of the cylinder.

    Let $h$ be the height of the cylinder, noting that $0 \lt h \lt 2r$ (as otherwise the cylinder won't fit inside the sphere). Then, the volume of the cylinder is given by $V(h) = \pi h (4r^2 - h^2) / 4$.

  22. A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus, the diameter of the semicircle is equal to the width of the rectangle.) If the perimeter of the window is $30$ ft, express the area of the window (a measure of how much light it admits) in terms of the radius of the semicircular part of the window.

    $\displaystyle{A(r) = 2r [ 15 - \frac{r}{2} (\pi + 2) ] + \frac{\pi r^2}{2}}$