BRIEF ANSWERS FOR TEST 1

• T or F?
  • T. Multiplying by 2 can change the range but not the domain.
  • F. For example, the domain of f(x) = square_root(x-2) is [2,infinity) while the domain of f(2x) is [1,infinity).
  • T. If y=f(x) is odd, then its graph is symmetric through the origin. The function y=-f(x), whose graph is obtained by reflecting the graph of y=f(x) across the x-axis, is also symmetric through the origin. Hence y=-f(x) is odd too.
  • F. Width is independent of length.
• Find the domain ...
  • (-3,infinity)
  • (-infinity,-1] union [1,4]
  • all real numbers except -4.
• Sketch ...
  • Start with the graph of the absolute value function, stretch vertically by a factor of 2, shift right two units, shift up two units.
  • For x less than or equal to 2 the graph should be a parabola with vertex at (1,-1) and which opens upward. This piece of the graph should join up with a piece of a second parabola defined for x greater than 2 which has vertex at (3,1) and which opens downward.
• f(x) is a piecewise defined function. It's definition is:
  (x+1)^2 for x less than or equal to 0
  1 for x greater than 0 but less than or equal to 1
  x otherwise
  
• Let the dimensions of the box be x, x, and y. Then the volume is V = x^2*y = 12, and the surface area is A = 2*x^2 + 4*x*y. Using the equation x^2*y = 12 we can eliminate x from A = 2*x^2 + 4*x*y. We get:
  A = 2*(12/y) + 4*(square_root(12/y))*y,
  which expresses A as a function of y.
• We want to maximize the area A = x*y given that x+2*y = 20. Thus we want to maximize A = (20-2*y)*y = -2*y^2 + 20*y = -2*(y-5)^2 + 50. The maximum area occurs when y = 5, and the maximum area is 50 square units.