Math 120R Summer Session II (2016) Quizzes

I will post quizzes and sketches of some of the solutions on this page.

Quiz 4 (15 July, 2016)

The following is a graph of some function f(x). The points A, B, C have coordinates (2,1), (3,4), (4,1), respectively. Sketch a graph of y = 1 - 3f(x-2) = g(x). What are the domain and the range?

quiz4_illustration.svgIdea: Note that we've only used the transformations we learned in class. The above graph shifts to the right by 2 units, stretches vertically by a factor of 3, and is reflected about the x-axis. It is then shifted up by 1 unit. Its asymptote becomes y = 1.

The domain of f is (-\infty, +\infty), and the range is (0,4]. The domain "shifts" by 2 units to the right, but since it's the whole real number, the domain of g remains (-\infty, +\infty). The range gets all the vertical transformations: it gets flipped and stretched by a factor of 3, then shifted by one unit up. Thus the range of g is [-11, 1) (note the boundaries!).

Another way to do this is to note that our new graph should look kind of like the original graph given above, so we can guess by evaluating g(4), g(5), g(6), then sketching the curve. (Note you still have to think a little bit about where the asymptote goes!)

Quiz 3 (14 July, 2016)

Let f(x) = \frac{x^{900} + 4x^{452} + x^2}{x^{9/2} - 4x^{5/2} + 4x^{1/2}}. Define g(x) = 3x^2 - f(a).

a) What is the independent variable in the definition of g(x)?
b) What is the domain of g?
c) What is the x-intercept?
c) For x>0, is this function increasing, decrease, or neither? Why?

Answer: a) x
b) (-\infty, +\infty)
c) \sqrt{f(a)/3}
d) Increasing

Idea: Since the independent variable is x, we can treat g(a) as a number! This means that f is such a parabola with a vertex at (0, -f(a)), and the answers follow:

a) The independent variable is x.
b) The domain of a quadratic is (-\infty, +\infty).
c) We solve 3x^2 - f(a) = 0 to get x = \pm \sqrt{f(a)/3}.
d) An upward-facing parabola is increasing to the right of its vertex.

Quiz 2 (13 July, 2016)

Find the domain of f(x), where

f(x) = \frac{x^{900} + 4x^{452} + x^2}{x^{9/2} - 4x^{5/2} + 4x^{1/2}}.

Answer: (0, \sqrt{2}) \cup (\sqrt{2}, +\infty).

Main idea: The numerator is just a polynomial, so it's defined everywhere. The denominator is only defined for x > 0 (since x^{1/2} = \sqrt{x} is only defined for x \geq 0). So is the domain just [0, +\infty)? No! We cannot divide by zero!

Using the fact that the denominator is equal to 0 when x = 0, \sqrt{2} (by the previous quiz), we arrive at the answer.

Quiz 1 (12 July, 2016)

Let f(x) = x^{9/2} - 4x^{5/2} + 4x^{1/2}. Solve the equation  f(x) = 0 for x.

Answer: x = 0 and x = \sqrt{2}.

Main ideas: Note the domain for f is x \geq 0, so any solution must be greater than or equal to 0.

Solve x^{9/2} - 4x^{5/2} + 4x^{1/2} = 0 by factoring out x^{1/2}:

x^{1/2}(x^4 - 4x^2 + 4) = 0.


x^{1/2}(x^2 - 2)^2 = 0.

The possible solutions are therefore x = 0, \sqrt{2}, -\sqrt{2}. The last possible solution is not in the domain of f, so we discard it.